What measurement is dependent upon the length of the lever arm and the angle between the force application and the lever arm?

Have you ever tried to pull a stubborn nail out of a board or develop your forearm muscles by lifting weights? Both these activities involve using a "lever-type" action to produce a turning effect or torque through the application of a force. The same torque can be produced by applying a small force at a larger distance (with more leverage) or by applying a larger force closer to the point about which the object has to rotate. These two examples are shown in Fig. 1. In the case of the hammer pulling the nail, a small force applied at the end of the handle translates into a larger force being exerted on the nail at a smaller distance from the point where the nail is fixed to the board. In the second example the weight on the palm of the hand is at a greater distance from the elbow. This requires the muscles to apply a larger force at a smaller distance, usually less than 5 cm from the elbow. These are both examples of lever action—force applied at a distance from a fulcrum or pivot point or axis of rotation. A force applied as described in the above examples results in a torque on a body. Torque usually produces a rotation of a body.

Figure 1: Two examples of torque

Torque is a measure of the turning effect of an applied force on an object, and is the rotational analogue to force. In translational motion, a net force causes an object to accelerate, while in rotational motion, a net torque causes an object to increase or decrease its rate of rotation. Torque is the product of the applied force and the perpendicular distance from the pivot point to the line of action of the force and is measured in units of N·m. where is sometimes called the "lever arm."

Consider the irregularly shaped two-dimensional object shown in Fig. 2a.

Figure 2: Illustration of lever-arm concept

• 1

  Sketch a line through the force. This dashed line in Fig. 2a represents the line of action of the force.

• 2

  Draw a perpendicular line from the axis of rotation O to the line of action of the force. This line, marked d in Fig. 2, represents the lever arm r defined in Eq. (1).

Figure 3: Dependence of lever arm on point of application of force

Since the lever arm d makes a right angle with the line of action of the force, the three quantities, d, F, and r make a right triangle. This triangle has been redrawn in Fig. 2b. From this triangle we see that the lever arm d is given by where is the angle between

r

F

Figure 4: A wheel experiencing two torques

By convention, torques causing counterclockwise rotations are considered to be positive and torques causing clockwise rotations are negative. In the above example

F

1

F

( 4 )

As mentioned above, torque is actually a vector. The torque vector is perpendicular to the plane formed by the vectors

r

F

Torque causes rotational motion with angular (or rotational) acceleration . where I is the moment of inertia of the system and is the angular acceleration. This equation is the angular equivalent of Newton's second law:

When the net torque is zero, the object will not change its state of rotational motion—i.e., it will not start rotating or stop rotating or change the direction of its rotation. It is said to be in rotational equilibrium. If the sum of the forces acting on the object is also zero, the object is in translational equilibrium and will not change its state of translational motion, that is, it will not speed up or slow down or change its direction of motion. Whenever both of these conditions

( 7 )

	τ = 0

	F = 0

• Meter stick
• Knife edge
• Known masses of varying values
• Unknown mass
• Balance

Procedure A: Balancing Torques

• 1

  Balance the meter stick on the knife edge. The point at which the stick balances is the center of gravity of the meter stick. Enter this value on the worksheet.

• 2

  Select two 200-gram masses and one 100-gram mass.

• 3

  Refer to Fig. 5 and Fig. 6. Place a hanger at the 20-cm mark, a distance cm to the left of the center of gravity and place mass on it.
• Place another hanger at the 65-cm mark, a distance cm to the right of the center of gravity and place a mass on it.
• Enter these values in Data Table 1.

Figure 5: Three balanced torques

Figure 6: Photo of experimental set-up

• 4

  Calculate the torques due to and enter these values in Data Table 1. Be sure to include the sign of the torques.

• 5

  Using the appropriate sign for each torque we can write the condition for rotational equilibrium as

( 8 )

	τ

• 6

  Use Eq. (8)

  	τ

  = m1gx1 − m2gx2 − m3gx3 = 0
  and the values of the torques due to to predict the torque due to (including its sign) and enter this value in Data Table 1. Be sure to include the sign of the torques.

• 7

  Use the predicted value of the torque due to to predict the position of at which the third mass must be placed to balance the meter stick. Enter this value on the worksheet.

• 8

  Experimentally determine the position and enter this value on the worksheet.

• 9

  Compare the two values for the position by finding the percent difference between the predicted and experimental values of See Appendix B.

Checkpoint 1:
Ask your TA to check your set-up and calculations.

Procedure B: Finding the Mass of a Meter Stick

• 10

  Move the knife edge to the 25-cm mark. You will notice that the meter stick is no longer in equilibrium. The unbalanced force is the weight of the meter stick acting at its center of gravity.

• 11

  Experimentally find the position, of the 200-gram mass, needed to balance the meter stick. Enter the value of on the worksheet.

• 12

  In the space provided on the worksheet, sketch and carefully label a diagram of the meter stick and the 200-gram mass. Show all the torque-producing forces. Remember that the weight of the meter stick acts at its center of gravity. Indicate on your diagram the directions (clockwise or counterclockwise) of each torque.

• 13

  Calculate the torque due to the 200-gram mass and enter this value in Data Table 2.

• 14

  Use the value of the torque due to the 200-gram mass and the conditions for rotational equilibrium to determine the torque due to the mass of the meter stick. Enter this value in Data Table 2.

• 15

  Using the value of the torque determined in step 14, calculate the value of the mass of the meter stick m2. Enter this value on the worksheet.

• 16

  Use the balance to measure the mass of the meter stick.

• 17

  Compare the measured and calculated values of the mass of the meter stick by computing the percent difference.

Checkpoint 2:
Ask your TA to check your diagram, set-up and calculations.

Procedure C: Determining an Unknown Mass

• 18

  Position the center of gravity of the meter stick over the support.

• 19

  Place a 50-gram mass at the 70-cm mark and a 200-gram mass at the 20-cm mark.

• 20

  You will tie the free end of the string to a shot bucket around the 1-cm mark and hang it over the pulley as shown in Fig. 7 and Fig. 8.

Figure 7: Set-up for determining an unknown mass

Figure 8: Photo of set-up for determining an unknown mass

• 21

  In the space provided on the worksheet, sketch and carefully label a diagram of this set-up. Show all the torque-producing forces. Indicate on your diagram the directions (clockwise or counterclockwise) of each torque.

• 22

  Calculate the torques due to and enter these values in Data Table 3.

• 23

  Use the values of the torques due to the two masses and the conditions for rotational equilibrium to determine the torque due to Enter this value in Data Table 3.

• 24

  Now add small masses to the bucket until the stick balances.

• 25

  Determine the mass of the shot and bucket using a balance.

• 26

  Compute the percent difference between the experimental and predicted values for the mass of the shot plus bucket.

Checkpoint 3:
Ask your TA to check your set-up, diagram and calculations.