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11/21/03
Chapter 5. Newton’s Laws of Motion
I. Newton’s Second Law of Motion
1. You are traveling in a car at 65 mi/h. At some point you see an obstacle ahead and
start braking until the car comes to a stop. Find the decelerating force exerted on
the car by the brakes if the mass of the car is 1.3 tons and your stopping distance
is 45 m. (Assume the deceleration is constant and there is no deceleration due to
friction).
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Comment:
First you must solve for the acceleration of the car using the equation for constant
acceleration. The problem gives v0, v, and x, so you can use the following
equation to solve for a: v2 = v02 + 2ax. Once you have acceleration, simply
substitute the appropriate values into the equation ΣF = ma. Note that mass m,
must be in kg.
2. A constant force of 20 N acts on a 12-kg body (object). If the body starts from
rest, how much time will it take the body to get up to speed of 4 m/s?
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Comment:
First you must solve for the acceleration of the $body using Newton's second law
of motion a = ΣF/m. Then, you know a, v0, and v, so you can use the following
equation to solve for t: v = v0 + at.
3. A particle is moving with zero acceleration while forces F1 and F2 are acting on
it. If F1  (5 N )i  (6 N ) j , what must be the magnitude of F2 ?
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Hint 2/Comment:
If acceleration is zero, consider what ΣF must be in order to make the equation ΣF
= ma true.
Comment:
If the acceleration of a particle is zero, the net force acting on that particle must
also be zero; thus, the forces F1 and F2 must sum to zero. So, the magnitude of F2
must be equal to that of F1, but they must be pointing in opposite directions.
Thus, use the Pythagorean theorem to find the magnitude of F1 from its
components, and this must also be the magnitude of F2.
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4. A free particle that has a mass of 10-19 kg is initially at rest. At some point it is
accelerated by a force. After the particle travels 10 m, it has a speed of 103 m/s.
What is the magnitude of the accelerating force?
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Comment:
In order to calculate the accelerating force, you must first find the acceleration of
the particle. The problem gives values of v0, x, and v, so you can use the
following equation to solve for acceleration, a: v2 = v02 + 2ax. Once you know
the acceleration, you can substitute the appropriate values into the equation ΣF =
ma to find the magnitude of the accelerating force.
5. A driver in a 1200-kg vehicle (car, truck), attempting to get through a snowdrift,
approaches it at a speed of 50 mi/h. The vehicle plows through 12 m of snow
before coming to a complete stop. What is the magnitude of the force exerted on
the vehicle by the snow?
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Comment:
In order to calculate the force exerted by the snow, you must first obtain
deceleration of the vehicle. The problem gives values for v, v0, and x, so you can
use the following equation to solve for acceleration, a: v2 = v02 + 2ax. Once you
know the acceleration, you can substitute the appropriate values into the equation
ΣF = ma to find the magnitude of the force exerted by the snow.
6. A 15-kg shopping cart is moving with constant velocity in the absence of friction.
At some point in time it reaches a 10o incline. Find the decelerating force that acts
on the cart while it is moving up the incline.
Hint 1/Comment:
The only forces acting on the shopping cart are gravitational force and the normal
force (the force exerted by the ground on the cart). Remember that only forces
applied in the direction opposite to the direction of motion cause deceleration.
Comment:
If you place the x-axis of the coordinate system along the incline, you will see that
the only force causing deceleration of the cart (the only force pointing opposite to
the direction of motion) is the horizontal component of the gravitational force.
Using trigonometric functions, you can solve for the magnitude of this force: F =
mg sin θ.
7. A shopping cart is moving with constant velocity of 5 m/s in the absence of
friction. At some point in time it reaches a 10o incline. Counting from the bottom
of the incline, what distance does the cart travel up the incline before it
momentarily comes to a stop?
Hint 1/Comment:
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Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Hint 2/Comment:
The only forces acting on the shopping cart are gravitational force and the normal
force (the force exerted by the ground on the cart). Remember that only forces
applied in the direction opposite to the direction of motion cause deceleration.
Hint 3/Comment:
First you must know the magnitude of the decelerating force in order to find the
magnitude of deceleration. This will then allow you to find the distance the cart
will travel before it momentarily comes to a stop.
Comment:
If you place the x-axis of the coordinate system along the incline, you will see that
the only force causing deceleration of the cart (the only force pointing opposite to
the direction of motion) is the horizontal component of the gravitational force.
Using trigonometric functions, you can find the magnitude of the decelerating
force to be equal to mg*sin θ. Then you can write down an expression for
Newton’s Second Law as follows: mg*sin ma, which yields to a =
g*sin(Use this value of acceleration in the equation v2 = v02 + 2ax to solve for
the distance the cart will travel.
8. A truck accelerates a 5,000-kg trailer from rest up to a speed of 35 mi/h over a
distance of 100 m. What is the force exerted on the truck by the trailer? Assume
that the acceleration of the trailer is constant.
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Hint 2/Comment:
By Newton's third law, force exerted by the truck on the trailer is equal to the
force that the trailer exerts on the truck.
Comment
In order to solve for the force exerted on the truck by the trailer, you must first
know the magnitude of acceleration. Since the problem gives values for v0, v, and
x, you can use the following equation to solve for acceleration a: v2 = v02 + 2ax.
After you obtain a, you can substitute the appropriate values into the equation ΣF
= ma to solve for the force.
9. A truck begins to pull a trailer from rest accelerating in up to 35 mi/h over 100 m.
What is the mass of the trailer if the force exerted by the truck on the trailer is
6500 N?
Hint 1/Comment:
Newton's second law of motion states that an object with mass m has an
acceleration a equal to the net force ΣF acting on that object divided by its mass
m: a = ΣF/m.
Comment:
In order to calculate the mass of the trailer, you must first determine the
magnitude of acceleration of the trailer. Since the problem gives values for v0, v,
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and x, you can use the following equation to solve for acceleration a: v2 = v02 +
2ax. After you obtain a, you can substitute the appropriate values into the
equation ΣF = ma to solve for the mass of the trailer.
II. Newton’s Third Law of Motion
1. An object (body, ball) is thrown against a wall. When it hits the wall, the object
exerts a force of 30 N on the wall. Calculate the force exerted on the object by the
wall.
Hint 1:
Newton’s third law of motion states that forces come in pairs. Those forces are
equal in magnitude but opposite in direction, or, more commonly, “for every
action there is an equal and opposite reaction.”
Comment:
By Newton’s third law, the force applied by the ball on the wall is equal in
magnitude to the force the wall applies on the ball, though the directions of these
forces are different.
2. Three boxes are placed side by side. A worker pushes the boxes with the force of
10 N directed parallel to the ground. Find the force that the box B exerts on box A.
(Find the force that box C exerts on box B) if the masses of the boxes A, B, and C
are equal 3.4 kg, 3.2 kg, and 5.0 kg respectively.
Hint 1:
Newton’s third law of motion states that forces come in pairs. Those forces are
equal in magnitude but opposite in direction or, more commonly, “for every
action there is an equal and opposite reaction”.
Hint 2:
To find the acceleration of one of the boxes (the common acceleration), treat all
three boxes as one large object and use their combined mass and the applied force
to solve for the acceleration.
Comment:
Use the common acceleration of the boxes and the mass of the box, on which the
force is being exerted, in the ΣF = ma equation to find the force exerted on that
box.
3. A 59-kg person driving a car accelerates up a 12o incline at 4.3 m/s2. Find the
magnitude of the force exerted on the driver by the car.
Hint 1:
If you place the coordinate system on the incline (with the x axis along the
incline), note that there is no acceleration in the y direction, so the normal force
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(the force exerted by the surface on the car) and the y component of gravity are
equal resulting in no net force in the y direction.
Hint 2:
The force exerted on the driver by the car equals to the net force acting on the
driver. According to Newton’s second law of motion, what is the net force acting
on a body equal to?
Comment:
According to Newton’s second law of motion, the net force acting on a body
equals to the product of the mass of the body and its acceleration (ΣF = ma).
4. A 60-kg person (scientist, physicist) is riding in an elevator that is accelerating
downward (upward) at 2 m/s2. What is the magnitude of the force exerted on the
elevator by the person?
Hint 1:
The force exerted on the elevator by the $person is equal in magnitude but
opposite to the force exerted on the $person by the elevator. The force exerted on
the $person by the elevator is also called the normal force.
Hint 2:
Suppose we direct the x axis upward. If the elevator is accelerating downward,
the accelerating force is pointing opposite to the normal force, reducing the net
force in the x direction. If the elevator is accelerating upward, the accelerating
force is pointing in the same direction as the normal force, increasing the net force
in x direction.
Comment:
The force exerted on the $person by the elevator can be found using Newton’s
second law of motion.
Keeping in mind that the axis is directed upward, we can write down the
equations projected on that axis for two cases.
Case of the elevator accelerating upward: ma = - mg + N.
Case of the elevator accelerating downward: - ma = - mg + N.
Finally, the magnitude of the force exerted on the elevator by the person is equal
to the magnitude of N.
5. On a crash test, a 1000-kg vehicle (car, truck, mini van) moving at 60 mi/h is
directed toward a high-density concrete wall. What is the magnitude of the force
exerted by the car on the wall if the impact took place over 0.1 s?
Hint 1:
Newton’s third law of motion states that forces come in pairs. Those forces are
equal in magnitude but opposite in direction or, more commonly, “for every
action there is an equal and opposite reaction.”
Hint 2:
In order to find the force exerted by the $vehicle on the wall, we need to find the
force exerted on the $vehicle by the wall, which is the stopping force applied to
the $vehicle. Use Newton’s second law of motion to find this force.
Comment:
According to Newton’s second law, the magnitude of the stopping force acting on
the $vehicle is expressed by the following equation: Fstopping = ma (assuming
there is no friction).
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Finally, the magnitude of the force exerted by the vehicle on the wall equals to
Fstopping.
III. Normal Forces
1. A block (brick) is being pulled by a force F as shown below. What is the possible
direction of the normal force vector N ?
a)
b)
c)
d)
e)
Comment:
A normal force is defined as a force exerted by a surface on a body. That force is
perpendicular to the surface.
2. A brick (block) is pushed against a wall by a force F as shown below. Assuming that
the brick is held in place, what is the direction of the normal force N ?
a)
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b)
c)
d)
e)
Comment:
A normal force is defined as a force exerted by a surface on a body. That force is
perpendicular to the surface.
3. Consider a block at rest on an incline. Which of the following pictures illustrates the
correct direction of the normal force N ?
a)
b)
c)
d)
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e)
Comment:
A normal force is defined as a force exerted by a surface on a body. That force is
perpendicular to the surface.
IV. Free-Body Diagrams
1. An overhead view of a 25-kg box being pulled by three people is shown below. Find
the magnitude of acceleration of the box if F1  11N , F2  21N , F3  5N , and if the
angles , , and are equal 70o, 45o, and 50o respectively. Assume there is no
friction.
Hint 1/Comment:
Approach this problem by applying Newton’s second law of motion to the box.
Using the data given in the text of the problem, we can easily find the net force acting
on the box in each direction. Because the box accelerates in two directions (x and y
directions), you will have two equations that contain ax and ay respectively.
Comment:
After you solve for ax and ay, the magnitude of acceleration can be found by using
Pythagorean Theorem.
2. Below is the diagram of a carriage pulled by two horses. Find the magnitude of the
force that each of the horses exerts on the carriage. The mass of the carriage is 350
kg, the acceleration is 3 m/s2, and  = 30o. Assume that there is no friction and that
both horses pull equally hard.
Hint 1:
Approach this problem by applying Newton’s second law of motion to the carriage.
Using the data given in the text of the problem, we can easily find the net force acting
on the carriage in each direction. Because the horses pull equally hard, the y
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components of the forces they exert will cancel each other out. Thus, the acceleration
in the y direction is equal to zero (ay = 0 m/s2).
Comment:
Finally, for the magnitude of the force F we get: F = max/(2cos).
3. A worker pushes a 50-kg cart up an incline with a force of 20 N directed horizontally.
Find the normal force acting on the cart (find the acceleration of the cart) if  = 20o?
Assume there is no friction.
Hint 1:
According to the way we positioned the coordinate system (see
picture), the cart does not accelerate in the y direction. Write down the equations for
Newton's second law along each of the axes. Then, having two equations and two
unknowns, you will be able to solve for the desired quantity.
Comment:
Newton's second law for the cart along the two axes is written as follows.
Along x: F (cos ) – mg (sin ) = max
Along y: N – mg (cos ) – F (sin ) = 0
Solve the system of equations for the desired quantity.
4. Two people are pushing a file cabinet as shown in the picture below. Find the
magnitude of acceleration of the file cabinet if F1  31N , F2  10 N ,  = 10o,  = 15o,
and the mass of the cabinet is 100 kg. Assume there is no friction.
Hint 1:
Because the file cabinet accelerates in both x and y directions, we first need to
find ax and ay (the x and y components of acceleration) and then use Pythagorean
theorem to find the magnitude of total acceleration.
Write down the equations for Newton's second law along each of the axes.
Comment:
According to Newton’s second law,
- max = - F1 cos  – F2 cos 
and
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– may = - F1 sin  – F2 sin .
Find a using Pythagorean Theorem (a2 = ax2 + ay2).
V. Other Applications of Newton’s Laws
1. An object placed on a scale is riding in an elevator. What is the measurement of
mass taken by an observer in the elevator if the mass of the object at rest is 2 kg?
Assume that the acceleration of the elevator is directed upward (downward) and is
equal to 3 m/s2.
Hint 1/Comment:
The reading of the scale will be equal to the force exerted on the scale by the
object divided by g, the acceleration due to gravity. According to Newton’s third
law, that force is equal in magnitude to the normal force acting on the object. We
can find the normal force by applying Newton’s second law of motion to the
object.
Hint 2/Comment:
Suppose we direct the x axis upward. If the elevator is accelerating downward,
the accelerating force acting on the object is pointing opposite to the normal force,
reducing the net force in the x direction. If the elevator is accelerating upward,
the accelerating force acting on the object is pointing in the same direction as the
normal force, increasing the net force in x direction.
Comment:
The force exerted on the object by the scale can be found using Newton’s second
law of motion.
Keeping in mind that the axis is directed upward, we can write down the
equations along that axis for two cases.
Case of the elevator accelerating upward: ma = - mg + N.
Case of the elevator accelerating downward: - ma = - mg + N.
Finally, the magnitude of the normal force acting on the object is equal to the
magnitude of N.
2. A smooth object (body) slides down a 35o incline. What distance does the object
travel in 5 s if it starts at rest? Assume there is no friction.
Hint 1/Comment:
Assume that the x axis is directed along the incline in the direction of motion of
the $object. Write down the expression for Newton's second law of motion with
respect to that axis and find the acceleration of the block.
Comment:
Having found the acceleration of the block, use the equation x = (1/2)at2 to find the
distance that the $object travels.
3. A robot picks up (lifts) a heavy rock (load) giving it an acceleration of 1 m/s2 and
exerting a force of 500 N. What is the weight of the rock?
Hint 1:
Write down Newton's second law for the $load and find the mass of the load.
Comment:
To find weight, multiply the mass of the $load by g, which is the acceleration due
to gravity.
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