Download 6.0 - Introduction 6.1 - Sample problem: a mass on

6.0 - Introduction
Now, you will get some additional practice applying Newton’s laws. More
specifically, you will use them in situations where multiple forces are acting on a
single object.
If the application of multiple forces results in a net force acting on an object, it
accelerates. On the other hand, if the forces acting on it sum to zero in every
dimension, the result is equilibrium. The object does not accelerate; it either
maintains a constant velocity, or remains stationary. (Forces can also cause an
object to rotate, but rotational motion is a later topic in mechanics.)
Equilibrium is an important topic in engineering. The school buildings you study in,
the bridges you travel across í all such structures require careful design to ensure
that they remain in equilibrium.
The simulation on the right will help you develop an understanding for how forces in
different directions combine when applied to an object. The 5.0 kg ball has two
forces acting on it, F1 and F2. They act on it as long as the ball is on the screen.
You control the direction and magnitude of each force. In the simulation, you set a
force vector's direction and magnitude by dragging its arrowhead; You will notice the angles are restricted to multiples of 90°. You can also
adjust the magnitude of each vector with a controller in the control panel. The net force is shown in the simulation; it is the vector sum of F1
and F2.
You can check the box “Display vectors head to tail” if you would like to see them graphically combined in that fashion. Press GO to start the
simulation and set the ball moving in response to the forces on it.
Here are some challenges for you. First, set the forces so that the ball does not move at all when you press GO. The individual forces must be
at least 10 newtons, so setting them both to zero is not an option!
Next, hit each of the three animated targets. The center of one is directly to the right of the ball and the center of another is at a 45° angle
above the horizontal from the ball. Set the individual vectors and press GO to hit the center of each target in turn.
The target to the left is at a 150° angle. It is the “extra credit” target. Determining the correct ratio of vectors will require a little thought. We
allow for rounding with this target; if you set one of the vectors to 10 N, you can solve the problem by setting the other one to the appropriate
closest integer value.
6.1 - Sample problem: a mass on ropes
The monkey hangs without moving in
the configuration you see here. What
is the magnitude of the tension in the
left rope?
Since it is stationary, the monkey is not accelerating, which means no net force is acting on it.
This section shows you a useful technique for solving problems that involve multiple forces acting on a single object. To calculate the overall
force on an object like the rope, the x and y components of each force need to be determined. Since there is no acceleration in any direction,
there is no net force along any dimension. Two equations can be developed: The sum of the x components equals zero, and the sum of the y
components equals zero.
We will use a consistent approach to solving multi-force problems. First we draw a free-body diagram to help us identify the variables and the
force components. Then we state the variables and their values when they are known. Next, we use Newton’s second law, relating the net
force to the acceleration and the mass of objects in the problem. Finally, we perform the algebraic and mathematical steps needed to solve the
problem.
Copyright 2000-2010 Kinetic Books Co. Chapter 6
161
Draw a free-body diagram
Variables
Since some of the forces do not lie along either the x or y axis, we use trigonometry to calculate the components of each force. For
consistency, we do this even for the weight, which points straight down in the negative y direction.
x component
y component
tension to left
TL cos 152.0°
TL sin 152.0°
tension to right
TR cos 43.0°
TR sin 43.0°
weight
0N
mg sin 270°
weight
mg = 3.06 N
What is the strategy?
1.
Draw a free-body diagram of the forces on the monkey.
2.
Set the net force equal to zero in each dimension.
3.
This gives two equations with two unknowns. Use algebra to solve for the left tension force.
Physics principles and equations
Newton’s second law
ȈF = ma
When the acceleration is zero, the forces along every dimension must sum to zero.
Step-by-step solution
We first apply the equilibrium condition in the
x dimension to relate the tension forces in the two ropes.
Step
Reason
1.
ȈFx = 0
no acceleration in x dimension
2.
TL cos 152.0° + TR cos 43.0° = 0
add x components
3.
TL(í0.883) + TR(0.731) = 0
trigonometry
4.
TR = TL(1.21)
solve for TR
We have one relationship between the two tensions. Now we use equilibrium applied in the y dimension to develop a second equation. The y
components of the rope tensions balance the monkey’s weight.
Step
Reason
5.
ȈFy = 0
6.
TL sin 152.0° + TR sin 43.0° + (3.06 N) sin 270° = 0 add y components
7.
TL(0.469) + TR(0.682) = 3.06 N
no acceleration in the y dimension
trigonometry
We now have two equations in two unknowns. We substitute one equation into the other and solve for the tension in the left rope.
Step
Reason
8.
TL (0.469) + TL (1.21) (0.682) = 3.06 N substitute equation 4 into equation 7
9.
TL(1.29) = 3.06 N
10. TL = 2.37 N
162
simplify
division
Copyright 2000-2010 Kinetic Books Co. Chapter 6
6.2 - Interactive checkpoint: helium balloon
A pair of strings holds a helium
balloon in place. Dan and Rosa are
each holding onto one of the strings.
There is no wind, but the atmosphere
provides a vertical lift force FL on the
balloon. The balloon’s weight is
negligible. What is the magnitude of
the tension in Rosa’s string? What is
the magnitude of FL?
Answer:
TR =
N
FL =
N
6.3 - Sample problem: pulling up a scaffold
What is the magnitude of the tension
in the rope above the window
washer's hands?
If you have ever worked in a skyscraper, you have probably seen window washers raising and lowering themselves and their scaffolding. Let’s
say a particularly energetic window washer is accelerating herself upward, as shown above. You are asked to find the amount of tension in one
of the two ropes.
Ropes and pulleys are fairly common in mechanics problems. We will always assume that tension is transmitted with a change in direction but
no change in magnitude in a rope that goes around a pulley. You may often be told to draw these conclusions when a problem states that the
pulley is assumed to be frictionless and massless, or the rope is massless and does not stretch.
The drawing above shows the weight of the combination of the scaffolding and the window washer, which is 931 N. To apply Newton’s second
law, we need to compute the mass of the system (window washer plus scaffolding) from its weight.
Draw a free-body diagram
Variables
weight
ímg = í931 N
tension in left rope
T
tension in right rope
T
acceleration
a = 0.400 m/s2
mass
m = 931 N / 9.80 m/s2 = 95.0 kg
Copyright 2000-2010 Kinetic Books Co. Chapter 6
163
All the forces are in the vertical direction, which simplifies our work. The weight is directed down, which means we will treat it as a negative
quantity in our equations. We treat the upward tensions as positive quantities.
What is the strategy?
1.
Draw a free-body diagram of the forces on the scaffolding.
2.
Calculate the net force on the scaffolding. In this case, the forces are all vertical, so we only need to consider forces along the y axis.
3.
Use Newton’s second law to find the tension force.
Physics principles and equations
Newton’s second law
ȈF = ma
Step-by-step solution
Step
Reason
1.
ȈF = ma
Newton's second law
2.
T + T + (–mg) = ma
net force along y axis, down is negative
3.
2T = mg + ma
rearrange
enter values
4.
5.
T = 485 N
solve for T
6.4 - Sample problem: blocks and a pulley system
What is the acceleration of the block
on the frictionless table?
Two blocks, labeled A and B, are connected by a rope running over a pulley, as shown above. The rope and pulley are massless, the rope
does not stretch, and the pulley and the table are friction-free. You are asked to find the acceleration of block A, on the table. This requires two
free-body diagrams, one for each block.
The magnitudes of the tension forces exerted on each block by the rope are the same; the amount of tension does not vary within a rope.
Because the rope connects the blocks, they accelerate at the same rate (but in different directions).
In problems like this, you have to inspect the situation to assign the correct sign to the acceleration of each block. Since the acceleration of the
block on the table is to the right, we use the usual convention that it is positive. Since the acceleration of the falling block is downward, we treat
it as negative.
Draw a free-body diagram
164
Copyright 2000-2010 Kinetic Books Co. Chapter 6
Variables
Block A on table
x component
y component
tension
T
0N
weight
0N
ímAg
normal force
0N
FN
acceleration
a
0 m/s
mass
mA = 4.20 kg
Falling block B
tension
0N
T
weight
0N
ímBg
acceleration
0 m/s
ía
mass
mB = 5.70 kg
What is the strategy?
1.
Draw a free-body diagram for each block.
2.
Calculate the net force on each block. Block A moves only in the horizontal direction, so we can ignore the vertical forces on it. Block B
moves only vertically, and there are no horizontal forces on it.
3.
Use Newton’s second law for each block to find two expressions for the tension force in the rope, and set the expressions equal to find
the acceleration.
Physics principles and equations
Newton’s second law relates the net force and acceleration. Block A moves only horizontally, so we will consider only the x direction for it;
similarly, we consider only the y direction for block B.
ȈF = ma
Step-by-step solution
We start by considering block A, and use Newton's second law to find an equation that gives an expression for the tension force in the rope.
Step
Reason
1.
ȈFx = mAax
Newton's second law applied to A
2.
T = mAa
tension is net force on A
Then we apply Newton's second law to block B to find another expression for the tension. Since block B falls, we assign its acceleration a
negative value.
Step
Reason
3.
ȈFy = mBay
Newton's second law applied to B
4.
T + (–mBg) = mB(ía)
net force is sum of tension and weight
5.
T = mBg í mBa
solve for T
We set the two expressions for the tension equal. The rest is algebra.
Step
Reason
6.
mAa = mBg – mBa
set tensions equal, from 2 and 5
7.
a = (mBg)/(mA + mB)
solve for a
8.
9.
enter values
a = 5.64 m/s2
evaluate
The steps above determine the magnitude of the acceleration. Since the question asked for the acceleration of the block on the table, the full
answer is 5.64 m/s2 to the right.
Copyright 2000-2010 Kinetic Books Co. Chapter 6
165
6.5 - Interactive problem: mountain rescue
You are in peril! You are hiking solo, with a basket of food and supplies. You have
just reached a ledge and the basket is at the bottom, attached to a rope you use to
haul up supplies. A hungry grizzly bear has appeared and you need to pull the
basket up as quickly as possible.
The basket is attached to a rope that passes through a carabiner (a metal ring that
acts as a pulley through which the rope can run freely). You have to pull this rope
with a constant force to rescue your food. The rope leaves the carabiner at a 16.0°
angle from the horizontal. The carabiner is fastened to the rock wall with a chock, a
device that is wedged into a crack in the wall. The carabiner is frictionless and the
massless rope does not stretch.
The basket has a mass of 18.0 kg. A horizontal force on the chock that exceeds
175 N will pull it loose. (Climbers are trained to consider the effects of forces pulling
in different directions on gear like chocks.) You need to calculate the maximum
force you can apply to the rope to lift the basket up. If the force is too great, the
chock will pull loose. If it is too small, the bear will have time to reach the food.
Calculate the pulling force on the rope that will result in a horizontal force that just
pulls the chock loose. Then reduce the pulling force by 2 N, and you will succeed. Enter the force value you calculate to the nearest newton
and press GO to start the simulation. Press RESET to try again.
If you have difficulty solving this, draw a free-body diagram of the forces on the chock, and review the problems in previous sections that deal
with forces applied at angles.
6.6 - Sample problem: airplane at constant velocity
The plane flies at an angle of 60° and
is not accelerating. Its weight is
2.60×105 N and its engine thrust is
3.00×105 N. What is the amount of lift
force on its wings?
Above, you see a jet airplane flying through the air at an angle. It travels at a constant velocity; this means it is not accelerating and no net
force is acting on the plane along any dimension.
The forces acting on the plane are its weight, the force provided by its engine, called thrust, the drag force from air resistance, and the lift force
from the wings. The lift force acts perpendicular to the surface of the wings. How much lift force do the plane’s wings provide to keep it aloft?
Draw a free-body diagram
There are four forces on the plane: its weight, the engine thrust, air resistance, and the lift force of the wings. Orienting the axes so that the x
axis lies along the length of the airplane will align three of these forces (thrust, air resistance, and lift) along an axis.
With this orientation of the axes, the weight vector makes an angle ș with the y axis. The right-hand diagram above shows that the angle ș in
the free-body diagram is the same as the angle at which the plane flies.
166
Copyright 2000-2010 Kinetic Books Co. Chapter 6
Variables
x component
y component
weight
ímg sin ș
ímg cos ș
thrust
FT
0N
lift
0N
FL
air resistance
FR
weight
0N
mg =
2.60×105
N
thrust
FT =
3.00×105
N
flight angle
ș = 60.0°
What is the strategy?
1.
Draw a free-body diagram of the forces on the plane, rotating the axes so three of the forces lie along an axis.
2.
Calculate the net force on the plane along the axis of the lift force, and solve for the lift force.
Physics principles and equations
Newton’s second law
ȈF = ma
When the acceleration is zero, the forces along every dimension must sum to zero.
Step-by-step solution
The lift force acts in the y dimension. In the y column of the variables table, all the values are known except the lift force. So, we need only
apply equilibrium in the y dimension to solve this problem.
Step
Reason
1.
ȈFy = 0
no acceleration in y dimension
2.
FL + (ímg cos ș) = 0
lift and y component of weight
3.
FL = (2.60×105 N) cos 60.0°
rearrange and enter values
4.
FL = 1.30×105 N
evaluate
6.7 - Sample problem: an inclined plane and static friction
The Surfer Bob toy is just about to
slide. What is the coefficient of static
friction?
A classic physics lab exercise asks you to use a block on a plane to calculate the coefficient of static friction for two materials. You see that
configuration shown above, although instead of a block, we are using Surfer Bob. You are given Bob’s weight and the angle the plane makes
with the horizontal just before Bob begins to slide. From this information, you are asked to calculate the coefficient of static friction. Since Bob
is not accelerating, no net force is acting on him.
You may have performed a lab experiment like this at some point during your studies. You incline a plane until the force of gravity just
overcomes friction and causes a block on the plane to slide. You then incline it a little less until the plane is at the angle at which the force of
static friction balances the force of gravity down the plane. At this point, the static friction is at its maximum and you can calculate the
coefficient.
As you see below, we can solve this problem in fewer steps by rotating the axes. Two of the forces are acting along the inclined plane. By
rotating the axes so the x axis is parallel to the plane, we can reduce the amount of trigonometry required. The rotation means that two of the
forces act solely along an axis. If we did not do the rotation, each force we analyzed would have to be decomposed into its x and y
components with the use of sines and cosines in order for us to solve the problem. If you do not like this axis rotation “trick,” then you can
always solve the problem by keeping the axes horizontal and vertical and using components.
Copyright 2000-2010 Kinetic Books Co. Chapter 6
167
Draw a free-body diagram
Variables
After we rotate the axes through the angle ș, the component of the force of gravity pulling Bob down the plane, and the frictional force, both lie
along the x axis. The component of the weight pressing Bob into the plane and the normal force are on the y axis. Since two forces (the normal
force and the frictional force) act solely along an axis, we do not have to calculate their components, and this reduces the amount of work that
follows.
When the axes are rotated, correctly identifying the component vectors that result can be difficult. The components are stated below in the
variables table. Note that the x component, for instance, is calculated using the sine, not the cosine of the angle ș. The correct trigonometric
ratios and signs must be determined using a diagram like the one above.
x component
y component
weight
ímg sin ș
ímg cos ș
normal force
0N
FN
friction
fs
0N
weight
mg = 0.100 N
What is the strategy?
1.
Draw a free-body diagram of the forces acting on Surfer Bob.
2.
Bob is not moving, which means he is not accelerating, so the forces in each dimension sum to zero. Sum the forces in the x dimension
to get an equation involving the coefficient of static friction.
3.
Sum the forces in the y dimension to get an expression for the normal force. Use it to solve the equation previously developed for the
coefficient of static friction.
Physics principles and equations
We use the equation for maximum static friction
fs,max = µsFN
Newton’s second law
ȈF = ma
When the acceleration is zero, the forces along every dimension must sum to zero.
Step-by-step solution
First we consider the forces in the x dimension. Bob is in equilibrium, so these forces sum to zero.
Step
Reason
1.
ȈFx = 0
no acceleration in x dimension
2.
µsFN + (ímg sin ș) = 0
substitute friction and x component of weight
3.
µs = (mg sin ș) / FN
solve for µs
We now have an equation for the coefficient of static friction, but we do not have enough information to evaluate it. We can, however, use the
fact that there is also equilibrium in the y dimension to get more information.
Step
Reason
4.
ȈFy = 0
no acceleration in y dimension
5.
FN + (ímg cos ș) = 0
normal force and y component of weight
6.
FN = mg cos ș
solve for normal force
168
Copyright 2000-2010 Kinetic Books Co. Chapter 6
Now we substitute the expression for the normal force from step 6 into the equation of step 3 and find the coefficient of static friction.
Step
Reason
7.
µs = (mg sin ș) / (mg cos ș)
substitute step 6 into step 3
8.
µs = sin ș / cos ș
simplify
9.
µs = tan ș
trigonometric identity
10. µs = tan 35.0°
enter value
11. µs = 0.700
evaluate
Even though we were given the weight of the dude, it turns out the coefficient of static friction does not depend on weight, but solely on the
angle of the plane.
6.8 - Interactive problem: desert island survival
You are stranded on a sweltering desert island. In addition to being sweltering, the
island is sloped. Fortunately, you have found a car. What exactly you will do with it
remains something of a mystery, but it is better than nothing. Because the car’s
brakes are broken, you have rigged a pulley system with a hanging block to hold the
vehicle stationary under the only tree on the island.
You need to calculate a mass for the hanging block that will keep the car stationary.
If the hanging block is too heavy or too light, the car will crash into a rock or roll into
the water. The mass of the car is 1210 kg and the angle of the slope from the
horizontal is 6.60°. The rope does not stretch and the pulley is massless and
frictionless. There is no friction between the car and the ground.
Enter the mass of the block to the nearest kilogram and press GO. If the
acceleration is zero, the car will not move, and you will have succeeded. Press
RESET to try again. (You may find it more entertaining to fail in this situation, but do
try to succeed.)
If you have difficulty solving this, draw a free-body diagram of the forces on the car
with the axes rotated so that the x axis matches the slant of the island. The entire force supplied through the rope must then equal a
component of the weight.
6.9 - Sample problem: an Atwood machine
What is the magnitude of acceleration
of the blocks?
The illustration above shows two blocks connected by a rope passing over a pulley. Because the blocks partially counterbalance each other,
the force required to lift either of them is less than its weight. This type of system is called an Atwood machine. An application can be found in
elevators, where a massive block partly counterbalances the weight of the elevator car to reduce the force required from the motor that lifts the
car.
In this sample problem, you are asked to find the rate at which the blocks accelerate. As usual, the rope and pulley are massless, the rope
does not stretch, and the pulley has no friction. The rope exerts an equal tension force on both blocks. The blocks’ accelerations are equal in
magnitude but opposite in direction.
Copyright 2000-2010 Kinetic Books Co. Chapter 6
169
Draw a free-body diagram
Variables
Block L
tension
T
mass
mL = 14.0 kg
weight
ímLg
acceleration
a
Block R
tension
T
mass
mR = 18.0 kg
weight
ímRg
acceleration
ía
The negative sign for the acceleration of block R indicates that it is directed downward. Because we are using the same variable a for the
magnitude of the acceleration of the two blocks, we must assign signs to their vertical acceleration components that reflect their directions, as
we do in the variables table.
What is the strategy?
1.
Draw a free-body diagram for each block.
2.
Calculate the net force on each block. There are no horizontal forces on either block.
3.
Use Newton’s second law for each block to find two expressions for the tension force in the rope, and set the expressions equal to find
the acceleration.
Physics principles and equations
Since both blocks move only vertically, we apply Newton’s second law in the y dimension.
Step-by-step solution
We start by applying Newton's second law to the block on the left to find an equation for the tension force. All the forces are in the vertical
direction, so we only consider the y components.
Step
Reason
1.
ȈFy = mLay
Newton's second law, left block
2.
T + (ímLg) = mLa
net force is tension plus weight
3.
T = mLg + mLa
solve for T
Then we apply Newton's second law to the block on the right, in the vertical direction, and find a second equation for the tension force.
Step
Reason
4.
ȈFy = mRay
Newton's second law, right block
5.
T + (–mRg) = mR(ía)
net force is tension plus weight
6.
T = mRg í mRa
solve for T
170
Copyright 2000-2010 Kinetic Books Co. Chapter 6
We set the two expressions for tension equal and solve for the acceleration.
Step
7.
Reason
mLg + mLa = mRg í mRa
set tensions equal, from steps 3 and 6
8.
solve for a
9.
enter values
10. a = 1.23 m/s2
arithmetic
Equation 8 can be profitably analyzed by considering a couple of special cases. If mL equals zero, equation 8 states that the acceleration
equals g. This makes sense: The block on the right would be in free fall, since no force would be opposing the force of its weight. Also, if
mR = mL, the acceleration would be zero, since there would be no net force. If we let mR go to infinity, equation 8 states that the acceleration
equals g. This too makes sense. If mR is very much bigger than mL, then mL will hardly slow down mR as it falls in free fall.
6.10 - Interactive checkpoint: two blocks
A 15.8 kg block sits on a frictionless
horizontal table. The block is attached
to a horizontal string that goes over a
pulley and is connected to another
block that hangs freely. The string is
massless and does not stretch. The
acceleration of the block on the table
is 3.89 m/s2. What is the mass of the
hanging block?
Answer:
mh =
kg
6.11 - Interactive checkpoint: two “blocks” at an angle
Frances, a 53.5 kg woman, slides on
a frictionless, icy ramp that is inclined
at 30.0° to the horizontal. An
unstretchable rope connects her to
Andre, who has a mass of 73.2 kg
and is accelerating at the same rate,
but parallel to the vertical wall of the
ramp. What is the magnitude of their
acceleration?
Answer:
a=
m/s2
Copyright 2000-2010 Kinetic Books Co. Chapter 6
171
6.12 - Sample problem: weight in an accelerating elevator
What is the magnitude of the normal
force exerted by the scale?
You may be familiar with feeling heavier or lighter in an elevator as it changes speed. When the elevator above is not accelerating, the scale
reports the person’s actual weight. When the person (Kevin) and scale are not accelerating, the scale indicates the amount of the gravitational
force that is exerted on Kevin, which equals mg.
When the elevator accelerates upward, Kevin feels heavier and the scale reports a value larger than his actual weight. When they accelerate
downward, the scale reports a smaller value. Each of these values is an apparent weight. Kevin’s actual weight does not change; it is the
acceleration that causes the scale to report these apparent weights.
In the problem, you are asked to determine Kevin’s apparent weight when the elevator is accelerating upwards at 2.00 m/s. You can do this by
determining the normal force that the scale exerts on him. A scale shows the amount of the normal force it is exerting on the person standing
on it. The normal force can be determined by applying Newton’s second law; the other force acting on Kevin (gravity) and his acceleration are
both stated.
Draw a free-body diagram
Variables
actual weight
ímg = í784 N
normal force
FN
acceleration
a = 2.00 m/s2
mass
m = 784 N / 9.80 m/s2 = 80.0 kg
What is the strategy?
1.
Draw a free-body diagram of the forces on the person.
2.
Determine the net force on the person.
3.
Use Newton’s second law to find the normal force.
Physics principles and equations
Newton’s second law
ȈF = ma
Step-by-step solution
Step
Reason
1.
ȈFy = may
Newton's second law
2.
FN + (ímg) = ma
substitute forces
3.
FN = mg + ma
solve for normal force
4.
FN = 784 N + (80.0 kg)(2.00 m/s2)
enter values
5.
FN = 944 N
evaluate
Equation 3 above is worth considering. If the acceleration a is zero, the normal force is the person's true weight, as should be expected. If the
elevator is in free fall, the acceleration a equals íg, so the terms on the right cancel. That means in free fall the normal force is zero and the
172
Copyright 2000-2010 Kinetic Books Co. Chapter 6
scale exerts no force on the person (and vice versa).
6.13 - Interactive summary problem: lunar module landing
You are in training to be an astronaut. In one of your training simulations, your goal
is to land a lunar excursion module (LEM) on the surface of the Moon. This LEM
has two rocket engines attached to the center of the craft. The engines typically
point at angles of 22.0° in opposite directions from the negative y axis and they
generate equal amounts of thrust. This means that as they fire, they provide a
combined upward force that does not cause the rocket to move to the right or left.
In an “accident” intended to test your astronaut skills, the right rocket (labeled B) is
bent to a 39.0° angle. The illustration to the right shows these two angles. In spite
of this problem, you are still expected to guide the LEM straight down.
There are two levels in this exercise. In the first, the module is moving downward
toward the landing pad as the simulation starts. In this “emergency,” the amount of
thrust force from the left engine, labeled A, is jammed at 35,600 N.
You are allowed to set the amount of force from the right engine. You need to set
this amount so there is no net horizontal force. If you set engine B’s force this way,
you will land the LEM gently onto the landing pad. Small attitude rockets at the top
of the LEM will keep it from rotating, but it will drift left or right if you enter the wrong
values.
To accomplish your first mission, compute the force needed from the right engine,
set that value to the nearest 100 N in the simulation, and press GO to test your
answer. Press RESET to try again.
The second exercise is harder and is optional. In the second simulation, the LEM is
again moving downward toward the landing pad. You need to set the forces of both
engines to achieve a net acceleration directly upward of 4.12 m/s2. The mass of the
module is 8910 kg, and it does not change significantly as the engines burn fuel. As
an astronaut, you know that the rate of acceleration due to the Moon's gravity is
1.62 m/s2.
To solve this problem, first calculate the total vertical force that should be provided
by the engines to counter the Moon's gravity and provide a net upward acceleration
of 4.12 m/s2. This will give you one equation involving the two unknown engine
forces. The horizontal components of the misaligned engine forces must still
balance so that the LEM stays on course. This gives you another equation. Solve the two equations and enter the force values to the nearest
100 N, then press GO.
If you have difficulty with this exercise, review the earlier sections in this chapter dealing with forces that do not act solely along one axis.
6.14 - Summary
This chapter focuses on applying concepts in sample problems. The basic concepts are:
1.
An object that is stationary has zero acceleration and has no net force acting on it. An object moving at a constant velocity also has no
net force acting on it.
2.
Draw free-body diagrams. They help you analyze problems.
3.
Break forces into components and sum those components along each axis. If the object is not accelerating, the sum of the x
components equals zero, as does the sum of the y components.
4.
Once you have calculated the net force, use Newton’s second law. Divide the net force by the mass to calculate the acceleration. (You
may have to divide the weight by the acceleration of gravity to determine the mass.)
5.
If there are two unknowns, you will need to develop two equations that include those unknowns.
And practice as needed!
Copyright 2000-2010 Kinetic Books Co. Chapter 6
173
Chapter 6 Problems
Chapter Assumptions
Unless stated otherwise, assume that all pulleys, ropes, strings, wires and other connecting materials are massless, frictionless and
otherwise ideal.
Conceptual Problems
C.1
Only one non-zero force acts on an object. Can the object be and remain at rest? Explain.
Yes
No
C.2
If an object has no velocity, does that necessarily mean no forces are acting on it? Explain your answer.
C.3
A massless rope is hanging over a massless and frictionless pulley. A large bunch of bananas is tied to one end of the rope,
and a monkey of equal weight is clinging to the other end, at a lower height. The monkey decides he wants the bananas and
starts to climb his side of the rope. What happens to the bananas as the monkey climbs?
Yes
No
i. The bananas rise at the same rate and stay out of reach
ii. The bananas stay in place
iii. The bananas fall at the same speed at which the monkey climbs
C.4
A car is parked on a downhill section of a hill. What forces are acting on the car?
C.5
Imagine using a spring scale on the Moon, where the acceleration of gravity is less than on Earth. Would the scale display
different values on the Moon and Earth? Explain why.
Yes
No
Section Problems
Section 0 - Introduction
0.1
Use the simulation in the interactive problem in this section to answer the following questions. (a) If F1 is set to 12 N directly
to the left, what should F2 be set to so that the ball does not move when you press GO? (b) If F1 is set to 10 N directly to the
left, what should F2 be set to so that the ball hits the target directly to the right of the ball? (c) If F1 is set to 10 N straight up,
what should F2 be set to so that the ball hits the target that is up and to the right of the ball?
(a)
(b)
N,
i. Straight up
ii. Directly to the right
iii. Straight down
iv. Directly to the left
i. Any magnitude less than 10 N
,
ii. Exactly 10 N
iii. Any magnitude greater than 10 N
(c)
N,
i.
ii.
iii.
iv.
i.
ii.
iii.
iv.
Straight up
Directly to the right
Straight down
Directly to the left
Straight up
Directly to the right
Straight down
Directly to the left
Section 1 - Sample problem: a mass on ropes
1.1
A child sits still on a swing that is supported by
two chains. Each chain makes a 15.0° angle
with the vertical. The child's mass is 25.0 kg.
What is the tension in each chain? (Ignore the
mass of the seat and chains.)
N
174
Copyright 2000-2010 Kinetic Books Co. Chapter 6 Problems
1.2
A tightrope walker is practicing. She is balanced on a rope at the exact midpoint of the rope. The tension in the rope is
4000 N, and the mass of the tightrope walker is 51 kg. Consider the left side of the rope. What angle with the horizontal does
it make? State this as a positive number in degrees.
1.3
A painting of mass 3.20 kg hangs on a wall. Two thin pieces of wire, each 0.250 m long, connect the painting's center to two
hooks in the wall. The hooks are at the same height and are 0.330 m apart. When the painting hangs straight on the wall, how
much tension is in each piece of wire?
°
N
1.4
1.5
1.6
A rubber chicken of mass 0.850 kg is dangling
as shown. What is the tension in each string?
T1=
N
T2=
N
A 4.50 kg block of gruyere cheese is
suspended as shown. Rope 1 on the left has a
tension of 28.3 N; rope 2 on the right has a
tension of 40.5 N. (a) What is ș1, the angle
made by rope 1? (b) What is ș2? State each
angle as the positive angle "inside" the triangle
the ropes form.
You may wish to use at least one of the
following trigonometric identities.
(a)
°
(b)
°
A circus has identical quadruplet acrobats.
Each has a mass m. They stand on a
tightrope, evenly spaced from one end to the
other, each a distance L from the next. At
each end, the string forms an angle ș1 with the
flat ceiling. The center section of rope is
horizontal and parallel to the ceiling. The
remaining 2 segments form an angle ș2 with
the horizontal. T1 is the tension in the leftmost
section of tightrope, T2 is the tension in the
section adjacent to it, and T3 is the tension in
the horizontal segment. (a) Find the tension in each section of rope in terms of ș1, m, and g. What is the tension T3? (b) Find
the angle ș2 in terms of ș1. If ș1 is 5.10°, what is ș2? (c) Find the distance d between the end points in terms of L and ș1.
Note: For all three parts of this question, derive the equations as asked. Parts (a) and (b) ask specific questions so that a
computer can evaluate your response; (c) does not.
(a)
2mg/sin ș1
mg/sin ș1
2mg/tan ș1
mg
None of the above
(b)
°
(c) Submit answer on paper.
Copyright 2000-2010 Kinetic Books Co. Chapter 6 Problems
175
Section 3 - Sample problem: pulling up a scaffold
3.1
3.2
A cat is stuck in a tree. You are designated with the job to get it out, yet you do not want to climb the tree, because you may
get stuck as well. Instead you set up a pulley system. A rope (consider it massless) runs from the seat you sit on over an ideal
pulley and then to your hand. You pull on the loose end of the rope with a force of 348 N. You weigh 612 N and the seat you
sit on weighs 16.0 N. (a) What is your acceleration? (b) What force does the seat exert on you?
(a)
m/s 2
(b)
N
Your fellow climber is stuck in a crevasse and
you need to get him out using a pulley system
shown in the diagram. You are pulling your
friend, of mass M, out of the crevasse at
constant speed by applying a force F. The
pulleys are frictionless and have negligible
mass, as do the ropes. Find each tension: (a)
T1, (b) T2, (c) T3, (d) T4, (e) T5 and (f) the
magnitude of F. (Hint: Drawing a free body
diagram for each pulley will help.)
(a)
(b)
(c)
(d)
(e)
(f)
176
i.
ii.
iii.
iv.
v.
vi.
vii.
i.
ii.
iii.
iv.
v.
vi.
vii.
i.
ii.
iii.
iv.
v.
vi.
vii.
i.
ii.
iii.
iv.
v.
vi.
vii.
i.
ii.
iii.
iv.
v.
vi.
vii.
i.
ii.
iii.
iv.
v.
vi.
vii.
N
Mg
Mg/2
2Mg
3Mg/2
Mg/3
5Mg
Mg/5
N
Mg
Mg/2
2Mg
3Mg/2
Mg/3
5Mg
Mg/5
N
Mg
Mg/2
2Mg
3Mg/2
Mg/3
5Mg
Mg/5
N
Mg
Mg/2
2Mg
3Mg/2
Mg/3
5Mg
Mg/5
N
Mg
Mg/2
2Mg
3Mg/2
Mg/3
5Mg
Mg/5
N
Mg
Mg/2
2Mg
3Mg/2
Mg/3
5 Mg
Mg/5
Copyright 2000-2010 Kinetic Books Co. Chapter 6 Problems
Section 5 - Interactive problem: mountain rescue
5.1
Use the information given in the interactive problem in this section to answer the following question. What is the force
required to save the food from the bear? As in the interactive, subtract 2 N from your answer to ensure that the chock does
not come loose. Test your answer using the simulation.
N
Section 6 - Sample problem: airplane at constant velocity
6.1
A climber of mass 64.8 kg is rappelling down a
cliff, but has momentarily paused. She stands
with her feet pressed against the icy,
frictionless rock face and her body horizontal.
A rope of negligible mass is attached to her
near her waist, 1.04 m horizontally from the
rock face. There is 5.25 m of rope between
her waist and where the rope is attached to a
chock in the face of the vertical wall she is
descending. Calculate the tension in the rope.
N
Section 7 - Sample problem: an inclined plane and static friction
7.1
Two people are pushing their stalled car up a hill with an incline of 4.20°. They are pushing parallel to the surface of the hill.
The car's mass is 954 kg. What is the combined force they must apply to keep the car moving at a constant speed up the hill?
Ignore the forces of friction and air resistance.
N
7.2
A block sits on an inclined plane, which is slowly being raised. The block remains motionless until the angle the plane makes
with the horizontal is 22°. At this angle, the block begins to slide down the plane. What is the coefficient of static friction
between the plane and the block?
7.3
You decide to find the coefficient of static friction between a paper cup and an archaic paper textbook. Armed with a metric
ruler, you place the cup on the edge of the book, and raise the edge of the book until the cup starts to slide downwards. The
length of the book is 26.0 cm, and you need to raise one edge 8.00 cm before the paper cup starts sliding. If the cup has a
mass of 211 grams, what is the coefficient of static friction between the cup and the book?
7.4
A beaver is pulling a small branch of mass m up a hill at a constant velocity. The hill slope is at an angle ș above the
horizontal. The coefficient of friction is µk. State how much force he must exert on the branch as he pulls it parallel to the
slope in terms of the variables stated in this problem.
mg sin ș + µkcos ș
mg sin ș + mgµk cos ș
mg cos ș + mgµk sin ș
None of the above
7.5
A skier of mass 64 kg starts from rest at the top of an 18° slope, points her skis straight down the slope and lets gravity pull
her down the slope. The slope is 65 m long. Her speed at the bottom is 15 m/s. (a) Assuming zero air resistance, what is the
magnitude of the frictional force between her skis and the snow? (b) What is the coefficient of kinetic friction between her
skies and the snow?
(a)
N
(b)
7.6
A block of mass 15.0 kg sits on a plane that is inclined at a 37.0° angle to the horizontal. A 227 N force, pointing up the plane,
is applied to the block. The coefficient of kinetic friction is 0.500. What is the speed of the block after 2.00 s?
m/s
7.7
When in deep space, a block accelerates at 4.2 m/s2 when a 120 N force is applied to it. The coefficient of static friction for a
particular plane and this block is 0.31. What angle can this plane make with the horizontal before the object starts to slide?
°
Copyright 2000-2010 Kinetic Books Co. Chapter 6 Problems
177
Section 8 - Interactive problem: desert island survival
8.1
Use the information given in the interactive problem in this section to calculate the mass for the hanging block to keep the car
stationary. Test your answer using the simulation.
kg
Section 9 - Sample problem: an Atwood machine
9.1
A 10.4 kg block sits on a frictionless horizontal table. The block is attached to a horizontal ideal string that goes over an ideal
pulley and is connected to another identical 10.4 kg block that hangs freely. What is the acceleration of the block on the
table? State the acceleration as a positive quantity.
9.2
Two blocks are connected by an ideal string that passes over a massless, frictionless pulley. The two blocks hang freely. The
first block has a mass of 14.3 kg, and the second block weighs 98.0 N. Determine (a) the magnitude of the blocks'
acceleration, (b) the magnitude of the tension in the string.
m/s 2
9.3
(a)
m/s 2
(b)
N
At a construction site, two barrels, each of mass of 122 kg, are connected by a long, massless rope that passes over an ideal
pulley that is mounted high overhead. The red barrel is at ground level, and the blue barrel is 8.26 m off the ground, at the
third story of a building. Some careless bricklayers hold the blue barrel in place, load it with 200 kg of bricks, and then let it
drop. (a) Calculate the maximum speed of the red barrel. (b) What is the maximum height that the red barrel reaches?
Remember that it is still moving upward even after the blue barrel hits the ground and stops.
(a)
m/s
(b)
m
Section 11 - Interactive checkpoint: two “blocks” at an angle
11.1 You and your friend are sledding on two sides
of a triangle-shaped hill. On your side, the hill
slopes up at 30.0° from the horizontal; on your
friend's side, it slopes down at the same
angle. You do not want to climb up the hill, so
you tell your friend to thread a rope through an
ideal pulley that is conveniently atop the hill.
He connects the rope to his sled and tosses
the other end of the rope to you. The sleds on
the snow have a coefficient of kinetic friction,
µk, of 0.0500. The total mass of your friend
and his sled is 82.0 kg while you and your sled have a mass of 68.0 kg. (a) What is the magnitude of the acceleration of each
sled? (b) What is the tension in the rope?
(a)
m/s 2
(b)
N
Section 13 - Interactive summary problem: lunar module landing
13.1 Use the information given in the first interactive problem in this section to calculate the force required from the right engine to
land the lunar module. Test your answer using the simulation.
N
13.2 Use the information given in the second interactive problem in this section to calculate the forces required by (a) the right
engine and (b) the left engine to land the lunar module safely. Test your answer using the simulation.
178
(a)
N
(b)
N
Copyright 2000-2010 Kinetic Books Co. Chapter 6 Problems
Additional Problems
A.1
You apply a force at an angle ș below the horizontal on a free-sliding sofa. Determine an expression for the acceleration of
the sofa as a function of the force F you apply, the sofa's mass m, and ș.
(F sin ș)/m
Fm sin ș
F sin ș
(F cos ș)/m
A.2
A wire of negligible mass is used to hang a
2.15 kg picture. The ends of the wire are
attached to the top 2 corners of the picture.
Each end of the wire forms an angle of 15.0
degrees from the horizontal. What is the
tension in the wire?
N
A.3
A.4
A llama pulls on a sled with a force given by F = 2.54i + 5.26j. A horse pulls on the same sled with a force defined by
F = 4.25i í 6.12j. The friction-free sled has a mass of 325 kg, and the forces are in newtons. (a) What is the net force on the
sled? (b) What is the acceleration of the sled? (c) If the sled starts at rest, what is its velocity after 4.50 seconds?
(a)
i+
(b)
i+
jN
j m/s 2
(c)
i+
j m/s
A 0.0820 kg pair of fuzzy dice is attached to the rearview mirror of a car by a short string. The car accelerates at a constant
rate, and the fuzzy dice hang at an angle due to the car's acceleration. If the dice hang at 5.50° from the vertical as the car
accelerates, what is the acceleration of the car?
m/s 2
A.5
A child pulls on a toy locomotive of mass 0.979 kg with a force of 3.25 N to the right. The locomotive is connected to two train
cars by cables. Friction in the axles results in an effective coefficient of kinetic friction between the floor and the train which is
0.110. One car has a mass of 0.952 kg and the other has a mass of 0.419 kg. (a) What is the acceleration of the train? (b)
What is the tension in the cable between the locomotive and the car connected to the locomotive?
(a)
m/s 2
(b)
N
Copyright 2000-2010 Kinetic Books Co. Chapter 6 Problems
179