Torque - AP Physics 1
Card 1 of 104
Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located at the 60cm mark with a weight of 30kg. At what point in between the two masses must the string be attached in order to balance the system?
Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located at the 60cm mark with a weight of 30kg. At what point in between the two masses must the string be attached in order to balance the system?
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This problem deals with torque and equilibrium. Noting that the string is between the two masses we can use the torque equation of 
. We can use the equation 
to find the torque. Since force is perpendicular to the distance we can use the equation 
(sine of 90o is 1). Force presented in this situation is gravity, therefore F=mg, and using the variable x as a placement for the string we can find r.
x=43, thus the string is placed at the 43cm mark.
This problem deals with torque and equilibrium. Noting that the string is between the two masses we can use the torque equation of
x=43, thus the string is placed at the 43cm mark.
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A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from it's left end. If a force of 50N is applied on it's right end, how much force would needs to be applied to the left end?
A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from it's left end. If a force of 50N is applied on it's right end, how much force would needs to be applied to the left end?
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This a an example of rotational equilibrium involving torque. The formula for torque is 
, where 
is the angle that the force vector makes with the object in equilibrium and 
is the distance from the fulcrum to the point of the force vector. To achieve equilibrium, our torques must be equal.
Since the forces are applied perpendicular to the beam, 
becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.
Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be 100N.
This a an example of rotational equilibrium involving torque. The formula for torque is
Since the forces are applied perpendicular to the beam,
Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be 100N.
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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
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The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.
Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.
.
Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.
The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.
Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.
Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the torque on the pulley when the system is motionless?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the torque on the pulley when the system is motionless?
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The net torque on the pulley is zero. Remember that 
, assuming the force acts perpendicular to the radius. Because the pulley is symmetrical in this problem (meaning the r is the same) and the tension throughout the entire rope is the same (meaning F is the same), we know that the counterclockwise torque cancels out the clockwise torque, thus, the net torque is zero.
In the image below, T1 (due to the platform with the 4 0.5kg weights) = T2 (the 2kg mass).
The net torque on the pulley is zero. Remember that
In the image below, T1 (due to the platform with the 4 0.5kg weights) = T2 (the 2kg mass).
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Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.
What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?
Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.
What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?
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Torque is defined by the equation 
. Since both students will exert a downward force perpendicular to the length of the seesaw, 
. In our case, force is the force of gravity, given below, and 
is the distance from the center of the seesaw.
Since the torque must be zero in order for the seesaw to stay parallel (not move), the lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other.
Torque is defined by the equation
Since the torque must be zero in order for the seesaw to stay parallel (not move), the lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other.
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Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.
Imagine that the two students are sitting on the seesaw so that the torque is 
. Which of the following changes will alter the torque of the seesaw?
Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.
Imagine that the two students are sitting on the seesaw so that the torque is
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Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student.
Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student.
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One side of a seesaw carries a 
mass four meters from the fulcrum and a 
mass two meters from the fulcrum. To balance the seesaw, what mass should be placed nine meters from the fulcrum on the side opposite the first two masses?
One side of a seesaw carries a
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For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.
The total torque must be equal on both sides in order for the net torque to be zero.
Substitute the formula for torque into this equation.
Now we can use the given values to solve for the missing mass.
The acceleration form gravity cancels from each term.
For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.
The total torque must be equal on both sides in order for the net torque to be zero.
Substitute the formula for torque into this equation.
Now we can use the given values to solve for the missing mass.
The acceleration form gravity cancels from each term.
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An attraction at a science museum helps teach students about the power of torque. There is a long metal beam that has one pivot point. At one end of the bar hangs a full sedan, and on the other end is a rope at which students can pull down, raising the car off the ground.
The beam is 40 meters long and the pivot point is 5 meters from one end. A car of mass 500kg hangs from the short end of the beam. Neglecting the mass of the beam, what is the minimum mass of a student who can hang from the rope and begin to raise the car off the ground?
An attraction at a science museum helps teach students about the power of torque. There is a long metal beam that has one pivot point. At one end of the bar hangs a full sedan, and on the other end is a rope at which students can pull down, raising the car off the ground.
The beam is 40 meters long and the pivot point is 5 meters from one end. A car of mass 500kg hangs from the short end of the beam. Neglecting the mass of the beam, what is the minimum mass of a student who can hang from the rope and begin to raise the car off the ground?
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We are trying to find what force needs to be applied to the rope to result in a net of zero torque on the beam.
Torque applied by the car:
We can use this to find the mass of a student that will create the same amount of torque while hanging from the rope:
We are trying to find what force needs to be applied to the rope to result in a net of zero torque on the beam.
Torque applied by the car:
We can use this to find the mass of a student that will create the same amount of torque while hanging from the rope:
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A bolt connecting the main and rear frame of a mountain bike requires a torque of 
to tighten. If you are capable of applying 
of force to a wrench in any given direction, what is the minimum length of the wrench that will result in the required torque?
A bolt connecting the main and rear frame of a mountain bike requires a torque of
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The minimum length of the wrench will assume that the maximum force is applied at an angle of 
. Therefore, we can use the simplified expression for torque:
Here, 
is the length of the wrench.
Rearranging for length and plugging in our values, we get:
The minimum length of the wrench will assume that the maximum force is applied at an angle of
Here,
Rearranging for length and plugging in our values, we get:
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There is a 
weight 
to the left the center of a seesaw. What distance from the center on the right side of the seesaw should Bob sit so that the seesaw is balanced?
Bob's mass is 
There is a
Bob's mass is
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Torque is defined as 
. In this case, 
is zero because Bob and the weight are sitting directly on top of the seesaw; all of their weight is projected directly downward. Therefore, the torque that the weight applies is:
In order for the seesaw to balance, the torque applied by Bob must be equal to 
.
Torque is defined as
In order for the seesaw to balance, the torque applied by Bob must be equal to
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Marc, Paul, and David all apply forces to a pendulum consisting of a rigid rod. Marc applies a force 
a distance 
from the pivot. If David applies a force of 
a distance 
from the pivot in the same direction as Marc, how much force must Paul apply in the opposite direction a distance of 
from the pivot if he is to make the sum of the torques about the pivot equal zero? Assume all three apply forces perpendicular to the rod.
Marc, Paul, and David all apply forces to a pendulum consisting of a rigid rod. Marc applies a force
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First, we must recall the formula for torque, which is
is the distance from the pivot, called the moment arm. 
is the force, and 
is the angle relative to the normal of the object.) Since all the forces are being applied perpendicular to the surface of the rod, 
. Thus,
The sum of the torques must equal zero, so David's torque plus Marc's torque must be the same as Paul's torque because David and Marc are applying torques in the opposite direction as Paul. This gives us
Dividing both sides by 
, we get Paul's force 
to be 
First, we must recall the formula for torque, which is
The sum of the torques must equal zero, so David's torque plus Marc's torque must be the same as Paul's torque because David and Marc are applying torques in the opposite direction as Paul. This gives us
Dividing both sides by
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Terry is pushing a vertical lever that is attached to the floor, and he pushes 
above the point of rotation. If he pushes with a force of 
at an angle of 
from the ground, what is the magnitude of torque that he is applying to the lever's hinge?
Terry is pushing a vertical lever that is attached to the floor, and he pushes
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Magnitude of torque can be found by relating the amount of force applied perpendicular to a lever arm about a point of rotation.
In this case, the force is not perpendicular, so we must take the perpendicular aspect of the force to find torque.
Plug in and solve.
Magnitude of torque can be found by relating the amount of force applied perpendicular to a lever arm about a point of rotation.
In this case, the force is not perpendicular, so we must take the perpendicular aspect of the force to find torque.
Plug in and solve.
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You apply a force of 
to a wrench of length 
. Determine the torque experienced by the bolt on the other end. Assume the force you apply is perpendicular to the wrench.
You apply a force of
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The definition of torque is
Where
is the force
is the distance
Theta is the angle between the direction of the force and the distance.
In this case, 
, so 
.
Plugging in our remaining values:
The definition of torque is
Where
Theta is the angle between the direction of the force and the distance.
In this case,
Plugging in our remaining values:
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From the given force and position vector, calculate the torque experienced by an object.
From the given force and position vector, calculate the torque experienced by an object.
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To calculate the torque experienced by the object, we must take the cross product of the force vector and the position vector.
To calculate the torque experienced by the object, we must take the cross product of the force vector and the position vector.
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Deanna and Rishi are playing on a seesaw (
long). Deanna (
) is 
to the right from the center of the seesaw (called a fulcrum) and Rishi (
) is on the left (see diagram). They want to try and see if they can balance horizontally (without teetering up or down). If Deanna doesn't want to move from her seat, where should Rishi move to, with respect to the fulcrum, in order to find this balance?
Deanna and Rishi are playing on a seesaw (
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This question deals with torques: 
It's important to note that torques are not (+) or (-), they are a direction, usually denoted as clockwise (CW) or counter clockwise (CCW). If you drew a FBD for each Deanna and Rishi, and included the pivot point, we'd see that each is in an opposite direction. That means that these torques oppose each other. If we want the seesaw to remain still, we must have the same torque on either side, this will result in a ZERO net torque. With that said, let's set this up!
The 
on either side can cancel eachother out. Now let's solve for the distance Rishi is from the fulcrum (
)
This question deals with torques:
It's important to note that torques are not (+) or (-), they are a direction, usually denoted as clockwise (CW) or counter clockwise (CCW). If you drew a FBD for each Deanna and Rishi, and included the pivot point, we'd see that each is in an opposite direction. That means that these torques oppose each other. If we want the seesaw to remain still, we must have the same torque on either side, this will result in a ZERO net torque. With that said, let's set this up!
The
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A symmetrical rectangle (
, 
) has four forces, all of the same magnitude, pulling at each corner as shown in the picture. Which of the following statements is true?
A symmetrical rectangle (
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First let's go on and define each of these terms...
Force: influence exerted
Torque: a way for us to measure of the effectiveness of a force which consists of a force and its perpendicular distance from the line of action amongst the axis of rotation
Each reference in which you find these definition may vary, but they should all have a common gist. Force is some type of implication on an object that can cause that object's mobility. Torque, however, is the capability of a specific force to create rotation. Therefore, torque is not a force, it's a characteristic of a force.
The forces shown in the diagram each have a reciprocal. In other words, 
is countered by
, as
is countered by 
. They each are equal, pulling in an opposite direction. With that said, the net force is zero. How? Well if we had a magazine (rectangle) and you and three of your friends each pulled on a corner as shown in the picture, the magazine wouldn't move. If you and a friend pulled on the same corners as
and 
,the magazine still wouldn't move (same goes for
and 
).
However, torques are slightly different. Torque is not (+) or (-), it is measured as clockwise or counter clockwise. So, looking at our diagram, all of the forces are in the same direction around the central pivot point (black circle). If three out of four of the forces shown were taken away, any of the remaining would cause the rectangle to rotate in the same manner (counter clockwise) around the pivot point. Therefore the net torque, or the total of the torques are all in the same direction and will NOT have a net value of zero, but rather a grand total of each force's torque separately.
First let's go on and define each of these terms...
Force: influence exerted
Torque: a way for us to measure of the effectiveness of a force which consists of a force and its perpendicular distance from the line of action amongst the axis of rotation
Each reference in which you find these definition may vary, but they should all have a common gist. Force is some type of implication on an object that can cause that object's mobility. Torque, however, is the capability of a specific force to create rotation. Therefore, torque is not a force, it's a characteristic of a force.
The forces shown in the diagram each have a reciprocal. In other words,
However, torques are slightly different. Torque is not (+) or (-), it is measured as clockwise or counter clockwise. So, looking at our diagram, all of the forces are in the same direction around the central pivot point (black circle). If three out of four of the forces shown were taken away, any of the remaining would cause the rectangle to rotate in the same manner (counter clockwise) around the pivot point. Therefore the net torque, or the total of the torques are all in the same direction and will NOT have a net value of zero, but rather a grand total of each force's torque separately.
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Michael just created a large square pinwheel. He attaches his pinwheel to a pivot screw in the middle of the square (side = 
) allowing it to spin in the wind. However, when he brings it outside, the pinwheel doesn't move. Frustrated, Michael gives the bottom corner a flick (
). What is the torque of the force provided?
Michael just created a large square pinwheel. He attaches his pinwheel to a pivot screw in the middle of the square (side =
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It is important to remember that torque is not a force, it's a characteristic of a force.
F is the force applied, r is the distance from the force's contact point and the object's center of mass, 
is the angle between 
and 
.
The trick to this problem is in two steps:
-
Find the 
using the Pythagorean theorem.
-
What's our 
? Well we know that every square has 4 
angles. If we drew a line from the corner of the square to the very middle (pivot point) of the pinwheel, we've just cut one of those angles in half! 
Solving:
Knowing our trig functions we can plug in for 
It is important to remember that torque is not a force, it's a characteristic of a force.
F is the force applied, r is the distance from the force's contact point and the object's center of mass,
The trick to this problem is in two steps:
-
Find the
-
What's our
Solving:
Knowing our trig functions we can plug in for
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Susie is trying to tighten a bolt (
) using a wrench (
, 
) as shown. Susie applies a force, 
, of 
. What is the net torque of the system? Note: the green "X" on the figure refers to the wrench's center of mass.
Susie is trying to tighten a bolt (
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Remember that the equation for torque is:
where 
is the force applied, 
is the distance from the force's contact point to the object's (wrench's) COM and 
is the angle between 
and 
We are provided with a whole bunch of numbers in the equation, however the explanation is much simpler than that. Looking at the direction of the force in reference to the COM, 
is zero.
Knowing our trig functions, we know that 
(regardless if we use radians or degrees).
The answer is 
Remember that the equation for torque is:
where
We are provided with a whole bunch of numbers in the equation, however the explanation is much simpler than that. Looking at the direction of the force in reference to the COM,
Knowing our trig functions, we know that
The answer is
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A man is tightening a bolt with a wrench. At what angle (with the wrench being the horizontal axis) and at what distance from the bolt should the man push for maximum torque?
A man is tightening a bolt with a wrench. At what angle (with the wrench being the horizontal axis) and at what distance from the bolt should the man push for maximum torque?
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The equation for torque is 
. Looking at this equation we can infer that the maximum distance from the center would give maximum torque. Also, any angle besides 
or 
will give an absolute value of a number below one, giving us a smaller torque.
The equation for torque is
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A force 
is applied to the edge opposite the doorhinge of a door of radius 
perpendicular to the door to produce a torque 
. Suppose now that the force is doubled, but now acts at a point 
from the doorhinge at an angle of 
to the door.
What is the resulting torque in terms of 
?
A force
What is the resulting torque in terms of
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The torque 
is produced by a force 
acting at a radius 
. Since the force and the radius are perpendicular, then the torque equation gives us:
The new torque, which we will call 
is produced by a force of 
acting at a radius of 
at an angle of 
. Thus the torque equation gives us:
Since 
, plugging this in to the above gives us 
The torque
The new torque, which we will call
Since
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