Download Skill Sheet 7.1A Adding Displacement Vectors

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Skill Sheet 7.1A
Adding Displacement Vectors
A displacement vector is a quantity that contains two separate pieces of information: (1) magnitude or size,
and (2) direction. When you add displacement vectors, you end up at a certain position. This new position is
the total displacement from the original position. A vector that connects the starting position with the final
position is called the resultant vector (x).
1. Example vector problem
Andreas walked 5 meters east away from a tree. Then, he walked 3 meters
north. Finally, he walked 1 meter west. Each of these three pathways is a
displacement vector. Use these displacement vectors to find Andreas’s total
displacement from the tree.
Displacement
vector
Direction
Magnitude
(meters)
Total magnitude (total
meters walked)
1
east
5
5
2
north
3
5+3=8
3
west
1
8+1=9
Andreas’s motion can be represented on a graph. To determine his total
displacement from the tree, do the following:
1.
2.
Add the east and west displacement vectors. These are in the x-axis
direction on a graph.
Andreas’s walk = 5 m east + (− 1)m west = 4 m east
Add the north and south displacement vectors. These are in the y-axis
direction on a graph.
Andreas’s walk = 3 m north
Solution: Andreas walked a total of 9 meters. The total displacement is 4 meters east and 3 meters north. The
resultant vector (x) goes from the starting position to the final position of total displacement.
2. Adding displacement vectors
1.
What is the total displacement of a bee that flies 2 meters east, 5 meters north, and 3 meters east?
2.
What is the total displacement of an ant that walks 2 meters west, 3 meters south, 4 meters east, and 1 meter
north?
3.
A ball is kicked 10 meters north, 5 meters west, 15 meters south, 5 meters east, and 5 meters north. Find the
total displacement and the total distance it traveled.
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Skill Sheet 7.1A Adding Displacement Vectors
3. Adding displacement vectors using x-y coordinates
A resultant vector can be written using x-y coordinates on a graph. The original
position is the origin of a graph where the axes represent east-west and northsouth positions. For example, (2,3)m is a resultant vector with the following
components: 2 meters east and 3 meters north. A resultant vector, (-3,-1)m, has
components 3 meters west and 1 meter south. Use this information to solve the
following problems. Write your answers using x-y coordinates. The first one is
done for you.
1.
Add the following four vectors to find the resultant vector, xR:
x1 = (5,0)m, x2 = (0,–5)m, x3 = (3,0)m, x4 = (–7,0) m
Add the east-west components: 5 m east + 0 m + 3 m east + (− 7) m west =
1 m east
Add the north-south components: 0 m + (−5) m south + 0 m + 0 m = (−5) m south xR = (1,−5)m.
2.
Add the following three vectors to find the resultant vector, xR:
x1 = (–2,0)m, x2= (0,–5,)m, x3 = (3,0)m
3.
Add the following vectors to find the resultant vector. Plot the resultant
vector (xR) on the grid to the left:
x1 = (4,0)m, x2= (–1,2)m, x3 = (0,1)m
4.
Add the following three vectors to find the resultant vector, xR.
x1 = (5,3)m, x2 = (–5,0)m, x3 = (5,2)m
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Skill Sheet 7.1B
Vector Components
There are two ways to represent displacement vectors. One way uses x-y coordinates. In this skill sheet, you
will practice representing vectors that use polar coordinates. Polar coordinates indicate the magnitude of the
vector and its angle from the x-axis on a graph. For example, a displacement vector (2 m, 10°) is 2 meters in
magnitude and at a 10° angle from the x-axis on a graph. Using trigonometry, you can convert polar
coordinates into x-y coordinates.
1. Two ways to write displacement vectors
A displacement vector, x, of magnitude 10 meters that has a component in the
x-direction equal to 7.07 meters and a component in the y-direction equal to
7.07 meters can be represented as x = (7.07,7.07)m.
This same vector can be represented in polar coordinates as x = (10.0,45°)m.
Why is this way of representing the vector like x = (7.07,7.07)m?
If the displacement vector is the hypotenuse of a right triangle, then:
x-component: x = 10cos(45°)m = 10(0.707)m = 7.07 m
y-component: y = 10sin(45°)m = 10(0.707)m = 7.07 m
Note that the angle is measured counterclockwise from the x-axis.
2. Problems
Answer the following problems and show your work. The first problem is done for you.
2.
Find the x-y components of the displacement vector x = (7.0, 30°)m.
x-component = (7.0 meters)cos 30° = 6.1 meters
y-component = (7.0 meters)sin 30° = 3.5 meters
x = (6.1, 3.5)m
What are the x-y coordinates for the displacement vector x = (10, 60°)m? Graph this displacement vector.
3.
Find the x-y components of the velocity vector Z =(17, 90°)m/sec.
1.
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Skill Sheet 7.1B Vector Components
4.
What are the x-y components of the displacement vector x = (100, 360°)cm.
5.
Graph the vector Z =(– 7.07,7.07)m.
6.
Graph the vector x =(– 2,– 4)cm.
7.
Find the components of the velocity vector Z =(80, 180°)km/h.
8.
Find the x-y components of the acceleration vector x = (5, 225°)m/sec2.
9.
Find the x-y components of the velocity vector Z =(80, 300°)km/h.
10. Find the x-y components of the force vector T = (80, 135°)N.
11. Challenge problem: A robot starts from a certain point and moves west for a distance of 5 meters; it then
goes north for 5 meters, and then east for 10 meters. Assume that the positive x direction is east and the
positive y direction is north. The negative x direction is west and the negative y direction is south.
a. Find the x-y components of the final displacement vector for the robot.
b. Express the final displacement vector in polar coordinates. Use the formula below to find the angle of the
resultant vector. Use the Pythagorean theorem to find r, the length of the resultant vector.
y-- = sin θ
r
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Skill Sheet 7.1C
Pythagorean Theorem
When you know the x- and y- components of a vector, you can find its magnitude using the Pythagorean
theorem. This useful theorem states that a2 + b2 = c2, where a, b, and c are the lengths of the sides of any right
triangle. For example, suppose you need to know the distance represented by the displacement vector (4,3)m.
If you walked east 4 meters then north 3 meters, you would walk a total of 7 meters. This is a distance, but it
is not the distance specified by the vector, or the shortest way to go. The vector (4,3)m describes a single
straight line. The length of the line is 5 meters because 42 + 32 = 52.
The Pythagorean theorem can be used to help us calculate the magnitude of a vector once we know its
components along the x- and y- directions. Also, we can find one of the components of the vector if we know the
other component and the magnitude of the vector.
1. Example problem
A displacement vector x = (2,3)m has
these components:
2 meters in the x direction.
3 meters in the y direction.
What is the magnitude of the vector?
Using the Pythagorean theorem, a is the
component along the x direction and b is
the component along the y direction. The
magnitude of the vector is c. We can find
the magnitude by taking the square root of a2 + b2:
2
2
a +b =
2
c
2
2
(2 m) + (3 m) =
2
2
4m +9m =
2
c
c
2
13 m = 3.6 m = c
2. Solving problems
1.
Find the magnitude of the vector u = (3, 4).
2.
Find the magnitude of the vector Z = (–3, –4).
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2
Skill Sheet 7.1C Pythagorean Theorem
3.
Find the magnitude of the vector Z = (5, 0).
4.
Find the magnitude of the vector x = (12.00, 6.00)cm.
5.
A robot starts from a certain point and moves east for a distance of 5.0 meters, then goes north for
3.0 meters, and then turns west for 2.0 meters.
a. What are the x-y coordinates for the resultant vector?
b. What is the magnitude of the resultant vector for the robot?
6.
Add the vectors Z1 = (5,0), Z2 = (0,–3), and Z3 = (1,0), and find the magnitude of the resultant vector.
7.
Add the vectors Z1 = (–5,0), Z2 = (0,–2), and Z3 = (7,0), and find the magnitude of the resultant vector.
8.
A resultant vector has a magnitude of 25 meters. Its y component is –12 meters. What is its x component?
3. Challenge problems
1.
Express the resultant vector in problem 5, Part 2 in polar coordinates. Assume that the positive x direction is
from west to east and the positive y direction is from south to north.
2.
Add the vectors Z1 = (5,0), Z2 = (0,–5), and Z3 = (5,180°), and find the magnitude of the resultant vector.
3.
Add the vectors Z1 = (5,45°), Z2 = (0,–10), and Z3 = (1,180°), and find the magnitude of the resultant
vector.
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Skill Sheet 7.2
Projectile Motion
This is a review of projectile motion. The problems in the skill sheet will give you practice in solving
problems that involve projectile motion.
1. Projectile motion has vertical and horizontal components
Projectile motion has vertical and horizontal components. Gravity affects the
vertical motion of an object. When we drop a ball from a height, we know that its
speed increases as it falls. The increase in vertical speed is due to the acceleration
gravity, g = 9.8 m/sec2. So the vertical speed of the ball will increase by 9.8 m/sec
after each second. After the first second has passed, the speed will be 9.8 m/sec.
After the next second has passed, the speed will be 19.6 m/sec and so on.
The acceleration of gravity affects only the vertical component of the motion.
Horizontal motion is not affected by gravity. So if we neglect the friction from air,
when we throw an object horizontally, its initial horizontal speed will not change.
For example, if we throw a marble horizontally at a speed of 5 m/sec, the marble
will be 5 meters horizontally from our hand after one second, 10 meters after
2 seconds, and so forth.
2. Solving projectile motion problems
Solving projectile motion problems requires using equations. To solve these problems, follow the steps:
•
Read the problem carefully. You may want to diagram the problem to help you understand it.
•
List what you know from the problem and what you need to solve for.
•
Determine which equations for vertical motion or horizontal motion will help you solve the problem. You
may need more than one equation to solve the problem. Some important equations are listed below.
•
Solve the problem and check your work.
2
2
2
Pythagorean theorem
a +b = c
Horizontal motion
v ox = v 0 cos θ
Use this equation to calculate initial horizontal velocity when you know
an angle and magnitude of the initial velocity vector.
Vertical motion
v oy = v 0 sin θ
Use this equation to calculate initial vertical velocity when you know an
angle and magnitude of the initial velocity vector.
Horizontal distance
x = v ox t
This equation is a rearranged version of the speed equation: v = d/t.
Here, x represents d, distance.
Vertical velocity
v y = v oy – gt
1 2
y = v oy t – --- gt
2
v oy
t = -----g
Gravity (g) is included in these equations because vertical speed
accelerates due to gravity when an object is falling.
Vertical distance
The time to reach
maximum height
Use this equation to find the magnitude of a velocity vector, (a, b).
This equation is a rearranged version of acceleration = speed/time.
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Skill Sheet 7.2 Projectile Motion
3. Sample projectile motion problem
A ball is kicked with an initial total velocity (v0) of 10 m/sec at an angle of 60 degrees off the ground. The time
that it takes for the ball to reach the ground again is twice the time it takes for the ball to reach its maximum
height. Using this information, estimate how far the ball will go horizontally and the maximum height it will
reach.
The horizontal (or x) component of the ball's velocity is:
vox = v0cos(60°) = 10 m/sec × 0.5 = 5 m/sec
The vertical (or y) component of the ball’s velocity is:
v oy = v 0 sin ( 60° ) = 10 m/sec × 0.87 = 8.7 m/sec
The time it takes for the ball to reach its maximum height (t) is written below. The initial vertical velocity is
voy and g is the acceleration due to gravity.
v oy
t = -----g
The total time it takes for the ball to travel horizontally is twice this long:
v oy
8.7 m/sec
t = 2 ×  ------- = 2 ×  -------------------------------- = 1.8 sec
g
9.8 m/sec/sec
With this information, we are now able to answer the questions:
• What is the horizontal range of the ball?
•
What is the maximum height reached by the ball?
The horizontal range equals the speed times the time in the horizontal direction:
x = vox × t = 5 m/sec × 1.8 sec = 9 m
The maximum height—or vertical distance (y)—can be calculated using the formula below.
1 2
y = [ v oy t ] – --- gt
2
In this problem, the ball reaches its maximum height in half the time that the ball travels before reaching the
ground. This time has been calculated to be 1.8 seconds. Therefore, 1/2 times 1.8 seconds is used in the
equation below for vertical distance.
2
1
1
1
y = v oy × --- ( time to travel horizontally ) – --- g × --- ( time to travel horizontally )
2
2
2
1.8
1
2 1.8
y = 8.7 m/sec × ------- sec – --- 9.8 m/sec ------- sec
2
2
2
2
2
= 3.82 m
Skill Sheet 7.2 Projectile Motion
4. Solving problems
Solve the following problems. Show your work.
1.
A cat runs and jumps from one roof top to another which is 5 meters away and 3 meters below. Calculate the
minimum horizontal speed with which the cat must jump off the first roof in order to make it to the other.
a. What do you know?
b. What do you need to solve for?
c. What equation (s) will you use?
d. What is the solution to this problem?
2.
An object is thrown off a cliff with a horizontal speed of 10 m/sec and some unknown initial vertical
velocity. After 3 seconds the object hits the ground which is 30 meters below the cliff. Find the initial
vertical velocity and the total horizontal distance traveled by the object.
a. What do you know?
b. What do you need to solve for?
c. What equation (s) will you use?
d. What is the solution to this problem?
3.
If a marble is released from a height of 10 meters how long would it take for it to hit the ground?
4.
A ball is thrown vertically upwards with a speed of 5 m/sec. How long before the ball hits the ground?
(HINT: Consider that there will be time for the ball to go up and then fall back down.)
5.
A ski jumper competing for an Olympic gold medal wants to jump a horizontal distance of 135 meters. The
takeoff point of the ski jump is at a height of 25 meters. With what horizontal speed must he leave the jump?
6.
An object is launched at an angle of 45 degrees with a speed of 20 m/sec. Calculate the initial velocity
components, the time it takes to hit the ground, the range, and the maximum height it reaches.
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Skill Sheet 7.3A
Equilibrium in 2-D
Here you will solve problems that require you to determine the unknown forced needed for an object to be in
equilibrium.
1. Forces and equilibrium
Forces are represented by vectors. They have magnitude (the strength of the
force) and direction. When forces are applied to an object, it will move unless
all forces acting on the object add up to zero. In this case the object is in
equilibrium.
2. Equilibrium problems
Do the following equilibrium problems. The first one is done for you.
1.
2.
A force T = (10, 30°)N is applied on an object. Find the x and y components of the force required for the
object to be in equilibrium.
x component = 10cos(30° + 180°) = –10cos30° N = –8.66 N
y component: = 10sin(30° + 180°) = –10sin30° N = –5.0 N
Find the force required to counteract the force T = (–5,10) kN.
3.
A box weighing 10 kilograms is pulled along the floor with a force of 100 N. The coefficient of friction µ
between the floor and the box is 0.5. Calculate the force required for the box to be in equilibrium.
4.
The force T = (100,–45°)N is applied to an object. Find the x and y components of the force required for
equilibrium.
5.
Find the x and y components of a force needed to counteract the forces T1 = (100,315°)N,
T2 = (100,225°)N, T3 = (60,150°)N.
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Skill Sheet 7.3B
Inclined Planes
When an object is placed on an inclined plane, the weight force has a component parallel to the plane and a
component perpendicular to the plane. The parallel component is pulling the object down the plane. Another
force acting on the interface between the object and the place surface is due to friction. The friction force is
acting parallel to the plane and opposite to the direction of motion. In this skill sheet, you will solve problems
that involve objects moving on inclined planes.
1. Solving inclined plane problems
The angle of an inclined plane, θ, is measured from the horizontal.
The horizontal component of the force is mgsinθ where m equals the mass of
an object and g equals the acceleration of gravity (9.8 m/sec/sec).
The vertical component of the force is mgcosθ.
The acceleration of an object on the plane is equal to gsinθ.
The friction force (Ffriction) equals mgcosθ multiplied by the coefficient of
friction (µ).
Ffriction = µmgcosθ
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Skill Sheet 7.3B Inclined Planes
2. Example problems
1.
Calculate the components of weight of a 10-kilogram box on an inclined plane making an angle 30 degrees
with the horizontal.
2.
What is the acceleration of the box in problem 1 along the plane?
3.
A mass of 30 kilograms is placed on an inclined plane making an angle of 25 degrees with the horizontal.
Find the force parallel and perpendicular to the plane and the acceleration along the plane.
4.
A box weighing 10 kilograms is placed on an inclined plane whose coefficient of friction is 0.30. Calculate
the maximum inclination angle before the box begins to move down the plane.
5.
What is the horizontal component of the force acting on the box in problem 4 at the maximum angle?
6.
What is the vertical component of the force acting on the box in problem 4 at the maximum angle?
7.
What is the total force along the plane acting on the box in problem 4 at the maximum angle?
8.
When the angle of the plane of problem 4 is increased to 45 degrees, what is the acceleration parallel to the
plane? Use the equation below to help you solve this problem:
acceleration = g ( sin θ – µ cos θ )
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Skill Sheet 8.1
Circular Motion
You used degrees when you first learned how to measure angles. However, the degree is not the most
convenient unit for using angles to calculate angular speed. For the purpose of angular speed, the radian is a
better unit of angle. One radian equals 57.3 degrees (approximately). Radians are better for angular speed
because a radian is a ratio of two lengths, and it does not have any units in the sense that meters or seconds are
units. This skill sheet provides you with practice in using degrees, radians, and in calculating angular speed.
1. Working with degrees, radians, and angular speed
A full circle has 360 degrees, or 2π radians (π = 3.14).
Convert degrees to radians by multiplying
by π/180°.
Example: How many radians is 45°?
π
45° × ----------- = 0.79 radians
180°
Convert radians to degrees by multiplying
by 180°/π.
Example: How many degrees are represented by 3.5 radians?
180°
3.5 radians × ----------- = 200°
π
Angular speed (ω) is given in radians/second. The formula for angular speed
is shown in the graphic at right.
Linear speed is found by multiplying angular speed by the radius of the
object being considered. The formula for linear speed is: v = ωr.
2. Example problems
1.
Convert to radians:
a. 0 degrees
b. 10 degrees
c. 30 degrees
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Skill Sheet 8.1 Circular Motion
d. 45 degrees
e. 90 degrees
f. 180 degrees
g. 270 degrees
h. 360 degrees
2.
Convert to degrees: 1.047 radians
3.
A wheel is spinning with an angular speed of 15 radians/sec. What is the angular speed in revolutions per
minute?
4.
A bicycle with a front wheel that is 50 centimeters in diameter and a back wheel that is 74 cm in diameter is
moving along with a linear speed of 16 km/hour. Find the angular speed of the wheels in radians/sec and in
rpm (revolutions per minute).
5.
A wheel that has a radius of 1 meter makes four turns in 3 seconds. Find the angular speed and the linear
speed of this wheel.
6.
A bicycle wheel with a radius of 0.5 meters is rolling with an angular speed of 1.75 rad/sec.What is the linear
speed of the wheel?
7.
A wheel with a radius of 0.25 meters is rolling with an angular speed of 0.75 rad/sec. How far will the wheel
go in one minute?
8.
A ball with a radius of 1 centimeter starts rolling down a ramp. The acceleration of the ball is 2 meters/sec2.
a. What is the angular speed of the ball after 1 second? After 3 seconds?
b. Convert each answer in 8(a) to revolutions per second.
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Skill Sheet 8.3
Universal Gravitation
The law of universal gravitation allows you to calculate the gravitational force between two objects from their
masses and the distance between them. The law includes a value called the gravitational constant, or “G.” This
value is the same everywhere in the universe. Calculating the force between small objects like apples or huge
objects like planets, moons, and stars is possible using this law as you will see as you solve the problems in
this skill sheet.
1. What is the law of universal gravitation?
The force between two masses m1 and m2 that are separated by a distance r is given by:
So, when the masses m1 and m2 are given in kilograms and the distance r is given in meters, the force has the
unit of newtons. Remember that the distance r corresponds to the distance between the center of gravity of
the two objects. For example, the gravitational force between two spheres that are touching each other, each
with a radius of 0.3 meter that are touching each other and a mass of 1,000 kilograms, is given by:
F = 6.67 × 10
– 11
2
N-m ⁄ kg
2
1,000
kg × 1,000 kg------------------------------------------------= 0.000185 N
2
( 0.3 m + 0.3 m )
This corresponds to a weight of a mass equal to 18.9 milligrams.
2. Example problems
Answer the following problems. Write your answers using scientific notation.
1.
Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by
a distance of 1 meter.
2.
A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 × 106
meters. Find the mass of the moon.
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Skill Sheet 8.3 Universal Gravitation
3.
m
For m = 5.9742 × 1024 kilograms and r = 6.378 × 106 meters, what is the value given by this equation: G ----2 ?
r
a. Write the answer in the blank below. Simplify the units of the answer.
b. What does this number remind you of?
c. What real-life values do m and r correspond to?
4.
The distance between Earth and its moon is 3.84 × 108 meters. Earth’s mass is m = 5.9742 × 1024 kg and the
mass of the moon is 7.36 × 1022 kg. What is the force between Earth and the moon?
5.
A satellite is orbiting Earth at a distance of 35 kilometers. The satellite has a mass of 500 kilograms. What is
the force between the planet and the satellite?
6.
The mass of the sun is 1.99 × 1030 kilograms and its distance from Earth is 150 million kilometers
(150 × 109 meters). What is the gravitational force between the sun and Earth?
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Skill Sheet 9.1
Torque
In this skill sheet, you will practice solving problems that involve torque. Torque is an action that is created by
an applied force and causes an object to rotate. Any object that rotates has a torque associated with it.
1. What is torque?
Torque, τ, can be calculated by multiplying the applied force, F, by r. The value,
r, is the perpendicular distance between the point of rotation and the line of
action of the force (the line along which the force is applied.
τ = F×r
The unit of torque is newton-meter (N-m).
For many situations the distance r is also called the lever arm.
2. Example torque problem
When you use a wrench to release a rusted bolt, you apply a torque around the axis of the bolt. You might have
noticed that the longer the wrench, the easier it is to perform the task. The length of the wrench is related to the
lever arm. However if you just pull at the end of the wrench you know that there is no way to release the rusted
bolt. The reason is that by pulling you have made the lever arm equal to zero. Zero lever arm, zero torque and the
bolt will keep on rusting.
Since the force has magnitude and direction so does the torque. We talk about torque in the counterclockwise
(CCW) direction, which we call positive and torque in the clockwise (CW) direction which we call negative.
Let’s do some numbers. If you have a wrench of length r = 30 centimeters and you apply a force of 1,000 N at the
end and perpendicular (90°) to the wrench. Because the force is applied downward so the wrench rotates
clockwise around the bolt, the force is negative (-1,000 N). The resulting torque is:
τ =F × r = -1,000 N × 0.3 m = -300 N-m
If you now apply the same force at an angle of 45 degrees from vertical the resulting torque is:
τ =F cos45° × r = -1,000 N cos 45° × 0.3 m = -212 N-m
Although you applied the same force, you get less torque at 45°. If you wanted to create a torque of -300 N-m
while applying the force at a 45 degree angle, you would need to apply -1,414 N of force!
– 300 N-m - = – 1,414 N
F = --------------------------------------cos 45° × 0.3 m
A see-saw works based on the ideas of torque. As you know, the lighter person
(or a cat!) has to sit further out for the saw to be level. You know now that this
is so because the only way to make the torque of the heavy person equal to the
torque of the light person is to increase the lever arm of the light person.
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Skill Sheet 9.1 Torque
3. Solving problems
Solve the following problems. Show your work. The first
problem is done for you as an example.
1.
For an object to be in rotational equilibrium about a certain
point, the total torque about this point must be zero. For the
example shown in the figure calculate the magnitude and
direction of a force that must be applied at point B for
rotational equilibrium about point P.
Solution: First note that the 35 N force does not create any
torque about point P because this force passes through that point (lever arm = 0).
Let, the force at point B equal FB.
For rotational equilibrium, the following must be true: Torque on to the left of P = Torque to the right of P
FB × 0.25 m = (50 N × 0.1 m) – (75 N × 0.4 m)
25 N-m- = – 100 N
F B = –--------------------0.25 m
For rotational equilibrium, 100 N must be applied downward at point B.
2.
A 10-kilogram mass is suspended from the end of a beam that is 1.2 meters long.
The beam is attached to a wall. Find the magnitude and direction (clockwise or
counterclockwise) of the resulting torque at point B.
3.
Two masses m1and m2 are suspended on an ornament. The ornament is hung
from the ceiling at a point which is 10 centimeters from mass m1 and
30 centimeters from mass m2.
a. If m1 = 6 kg, what does m2 have to be for the ornament
to be in rotational equilibrium?
m
b. Calculate the ratio of -----1- so that the ornament will be
m2
horizontal.
c. Suppose m1 = 10 kg and m2 = 2 kg. You wish to place a third mass, m3 = 5 kg, on the ornament to make it
balance. Should m3 be placed to the right or to the left of the ornament’s suspension point? Explain your
answer.
d. Calculate the exact location where m3 should be placed.
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Skill Sheet 9.1 Torque
4.
Forces are applied on the beam as shown on the figure at right.
a. Find the torque about point P produced by each of the three forces.
b. Find the net torque about point P.
c. A fourth force is applied to the beam at a distance of 0.30 m to the right of point P. What must the
magnitude and direction of this force be to make the beam in rotational equilibrium?
5.
A flag pole 2.5 meters long is attached to a wall at a 40° angle from vertical. A 50-kilogram mass is
suspended at the end. Calculate the resulting torque at the point of attachment to the wall.
6.
Three 10-newton forces act on a beam as shown to the right.
a. Calculate the torque produced by force 1 about point P.
b. Calculate the torque produced by force 2 about point P.
c. Calculate the torque produced by force 3 about point P.
d. Calculate the net torque about point P.
3