LAB 5: TORQUE

There is a good reason for attaching a doorknob as far as possible from the hinges of the door. When you open a door, you must apply a force. Where you apply that force and in what direction you push or pull determine how easily the door will rotate open. Forces not only produce motion in a straight line, they also produce rotation in a rigid body that has an axis. The turning movement caused by one or more forces acting on a body that is free to rotate about an axis is known as torque, T, and is defined as the product of the vector quantity force, F, and lever arm, r. The lever arm is the perpendicular distance from the axis of rotation to a line drawn in the direction of the force. Thus, the farther this line is from the axis, the more effective is the force that causes rotation.

In this experiment, two parallel forces, the two spring scale balances, will balance the downward force of the hanging mass, as shown in Figure A. The two parallel forces will tend to cause rotation because each is exerted over a specific distance, the lever arm, from the third force, the weight of the hanging mass. The weight force is at the pivot, point B in the figure. This fixed point B is the axis of rotation.

The force at point A acts over a distance AB and produces a clockwise torque (a negative value). The force at point C acts over a distance BC and produces a counterclockvrise torque (a positive value). The sum of these two torques may cause the meteistick to rotate in either a clockwise or counterclockwise direction about the axis. If the sum of all the clockwise torque's about the axis and the sum of all the counterclockwise torque's about the same axis equal zero, the meterstick will not rotate. In this experiment, when the two parallel forces are balanced by the third force; the sum of all torques will equal zero, and the system will be in rotational equilibrium.

OBJECTIVES:

MATERIALS:

2 spring scales, 1 half meterstick, 3 meterstick clamps, 1 set of hanging masses

Procedure

1. Set up the apparatus as shown in Figure A, without the hanging mass. Hold the two spring balances from supports on the laboratory table so they hang over the edge of the table. Be sure that the spring scale mechanisms can move freely.

2. For the first trial, center one clamp 5-cm from one end of the meterstick (point A) and center the other clamp 5-cm from the other end of the meterstick (point C), as shown in the figure.
  1. Record the force readings on the spring scales.

4. Hang a mass (500 g if you have a Sargent-Welch scale, less if you have Ohaus scale) from the clamp at point B, at the center of the meterstick. Record the scale readings at points A and C when the apparatus is in equilibrium

5. Measure and record the distances AB and BC.

6. Repeat steps 3 - 5 for trials 2 through 5, but moving the clamp at points B (middle clamp) to different positions on the meterstick.

Hanging mass: _______

DATA TABLE:

trial

Length

AB

Force without weight

Force with weight

Length

BC

Force without weight

Force with weight

1

 

 

 

 

 

 

2

 

 

 

 

 

 

3

 

 

 

 

 

 

4

 

 

 

 

 

 

5

 

 

 

 

 

 

CALCULATIONS:

Do the following calculations for each of the five trials and record them on the result table.

  1. The actual force of each spring scale is the difference between the force with, and without the hanging mass. Calculate the actual force for each scale.
  2. Draw a free body diagram of the ruler. Show and label all three forces. Show the torque about point B due the each of the forces.
  3. The clockwise torque about point B is equal to the product of the actual force for spring scale at A and the distance AB. The counterclockwise torque about point B is equal to the product of the actual force for spring scale at C and the distance BC. Calculate the clockwise and counterclockwise torque's about B.
  4. Draw another free body diagram of the ruler. Show the torque about point A due the each of the forces.
  5. Repeat step 2 of the calculations but calculate the clockwise and counterclockwise torque's about point A instead. Remember that the lever arm is the distance from the pivot to the point of contact. Since you are looking for the torque about point A, point A is now your pivot.
  6. Calculate the net torque for each trial for both steps 3 and 5 of the calculations.
  7. Calculate the percent error of the net torque by dividing the net torque by the average of the magnitudes of the clockwise and counterclockwise torques for each trial.

SAMPLE CALCULATIONS:

RESULTS TABLE:

trial

Actual force at A

Actual force at B

(weight of the hanging mass)

Actual force at C

1

 

 

 

2

 

 

 

3

 

 

 

4

 

 

 

5

 

 

 

 

 

trial

Clockwise torque about A

Counter-clockwise torque about A

Clockwise torque about B

Counter-clockwise torque about B

1

 

 

 

 

2

 

 

 

 

3

 

 

 

 

4

 

 

 

 

5

 

 

 

 

trial

net torque about A

%

error

net torque about B

%

error

1

 

 

 

 

2

 

 

 

 

3

 

 

 

 

4

 

 

 

 

5

 

 

 

 

ANALYSIS:

  1. What should the net torque be for step 3 in the calculations and for step 5 in the calculations? Explain.
  2. How does the force on point A compare with the force on point C for each trial? Explain why.
  3. Explain the results of steps 6 and 7 in the calculations.
  4. What do you observe about the relationships between the forces and the lever arms?

CONCLUSION: Summarize your findings and thoughts about the lab.

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