Torques, and the Center of Gravity

Pre-lab questions

1. What are the conditions for an object to have (a) translational, (b) rotational equilibrium?

2. If the conditions of question 1 are met, is the object necessarily in static equilibrium?

3. What quantities must be known to solve for a torque in this lab?

I. INTRODUCTION

An object is in static equilibrium if it has no translational or rotational motion. For this to occur the net force acting on an object must be equal to zero and the net torque acting around any point or axis of rotation must be equal to zero. In this experiment a meter stick and suspended weights will be used as a "beam" balance to study torques and the center of gravity.

II. EQUIPMENT LIST

+ Meter stick

+ Support stand

+ String and knife-edge clamps

+ Four hooked weights (50, 100, 200 g) + Unknown mass with hook (or string) + Laboratory balance

III. BACKGROUND INFORMATION

An object is in mechanical equilibrium when the following conditions are met: F = 0 = 0

The sum of the forces F and the torques acting on the object are zero. When the sum of the forces equal zero the object is in translational equilibrium and the velocity of its center of mass will remain constant. In this experiment the meter stick will be restrained from moving so this first condition will be met and the velocity of the center of mass will be zero (static translational equilibrium).

An object is in static equilibrium when it is in translational equilibrium and rotational equilibrium. To be in rotational equilibrium the object's rotational motion must be constant. This is achieved when the net torque acting on the object about any point equals zero. If the object is at rest and it is in rotational equilibrium it will not begin to rotate. The magnitude of a torque is given by the product of the force and the lever arm. The lever arm is the perpendicular distance from the line of action of the force to the axis of rotation. We will be dealing with essentially two-dimensional objects in the lab and can consider rotations in a plane. The rotations caused by a torque will be either counterclockwise or clockwise. For the system to be in equilibrium the counterclockwise torques must equal the magnitude of the clockwise torques.

For an extended body the gravitational force on the differential elements of mass in the object supply torque to the object about an axis of rotation. The center of gravity is the point of the object about which the sum of the torques provided by the gravitational force is equal to zero. (Note: When the gravitational acceleration is a constant over the object the center of gravity is the same point as the center of mass.) By placing an appropriate size single upward force with its line of action passing through the center of gravity the object can be balanced.

IV. EXPERIMENTAL PROCEDURE

A. Determine the mass of the meter stick (without any clamps). Place a knife-edge clamp near the center of

the meter stick and place the meter stick on the support stand. Adjust the meter stick through the clamp until the stick is balanced on the stand. Tighten the screw of the clamp and record the position of clamp. Note if using weight hangers, their mass must be added to the total for solving for the net force and torque

B. Case 1: Suspend m1 = 100 g at the 10 cm position. Suspend m2 = 200 g at the appropriate position to balance the meter stick. Record the position and compare the torques. To solve for the torque of each force solve for the lever arm and use the weight of the total mass hanging from the meter stick.

C. Case 2: Suspend m1 = 100 g at the 25 cm position and m2 = 200 g at the 75 cm position. Suspend m3 = 50 g at the appropriate position to balance the meter stick. Record the position and compare the torques.

D. Case 3: Suspend m1 = 100 g at the 25 cm position and m2 = 200 g at the 75 cm position. Suspend m3 (the unknown mass) at the appropriate position to balance the meter stick. Record the position and compute the torques. Measure the mass of m3 on the laboratory balance.

E. In the previous case the meter stick was not taken into account for the calculations because the support was at the meter stick's center of gravity. In the following cases the support will not be at the center of gravity of the meter stick. The meter stick's weight will contribute a torque to the system and will be used in the calculations.

Case 4: Suspend m1 = 100 g near the zero end of the meter stick. Move the meter stick in the support clamp until the system is in equilibrium. Record the position of the clamp. Compute the torque for m1 with the lever arm as the distance between the mass and the support. Consider all the mass of the meter stick as acting at its center of gravity. Solve for the torque provided by m1 and then use the lever arm for the meter stick's mass to solve for the mass of the meter stick. Compare the value to that measured on the balance.

F. Case 5: Suspend m1 = 100 g as close to the 0 cm end of the meter stick as possible, suspend m2 = 100 g at the 65 cm point on the meter stick. Adjust the meter stick in the support clamp until the system is in balance. This locates the center of gravity of the system. Compute the location of the center of gravity of the system and compare the computed and experimental values.

G. Case 6: Repeat the situation in case 5 but with m2 at the 80 cm point.

V. DATA

Your data table must keep track of the masses (including the meter stick) and their respective positions, and the support position.

VI. CALCULATIONS

1. For cases 1 & 2; compute the weight, and torque of each mass. Compute and compare the total counterclockwise and clockwise torques. Compute the percentage difference.

2. For cases 3 & 4; compute the value of the unknown mass from the torques. Compute the percentage error.

3. For cases 5 & 6; compute the torques and find the position of the center of gravity. Compare to the

experimental value and compute the percentage difference.

VII. CONCLUSION/QUESTIONS

1. What are possible sources of errors in the experiment?

2. Explain why the weight of the support is not used in any of the torque calculations.

3. Explain the principle of the dual-beam laboratory balance using the idea of torques.

4. When suspending the meterstick by itself, it is possible that the support position for the meter stick might

not be at its geometric center. Explain.

Lab report

Pre-lab questions

1. A) In order for an object to have a translational equilibrium, the net force that acts on the object has to be zero is the condition that has to be met.

b) In order for an object to have a rotational equilibrium, the net torque of the object needs to also be zero

2. For an object to be in static equilibrium, it must have no acceleration even during a motion. Therefore, it is necessary to have net torque and net force be zero.

3. The quantities that are needed to find torque are r which is the distance in meters and F which is fore in Newtons. Together, the torque's formula is t (torque) = r * F

V. DATA

Mass of the ruler: 0.15720 kg

Mass of weight 1: 0.1003 kg ; Mass of clamp 1: 0.01632 kg Mass 1: 0.11662 kg

Mass of weight 2: 0.20021 kg ; Mass of clamp 2: 0.01652 kg Mass 2: 0.211862 kg

Mass of weight 3 (Case 2): 0.05020 kg ; Mass of clamp 3: 0.01643 kg Mass 3: 0.06663 kg

Mass of weight 3 (Case 3): 0.50035 kg ; Mass of clamp 3: 0.01643 kg Mass 3: 0.51678 kg

*Distance are measured from the center of gravity

Distance 1 (m)

Distance 2 (m)

Distance 3 (m)

Sketch

Case 1

0.402

0.228

N/A

Case 2

0.252

0.248

0.352

Case 3

0.252

0.248

0.044

Case 4

0.300

N/A

N/A

Case 5

0.390

0.260

N/A

Case 6

0.435

0.365

N/A

Mass of the ruler: 0.15720 kg

Mass of weight 1: 0.1003 kg ; Mass of clamp 1: 0.01632 kg Mass 1: 0.11662 kg

Mass of weight 2: 0.20021 kg ; Mass of clamp 2: 0.01652 kg Mass 2: 0.211862 kg

Mass of weight 3 (Case 2): 0.05020 kg ; Mass of clamp 3: 0.01643 kg Mass 3: 0.06663 kg

Mass of weight 3 (Case 3): 0.50035 kg ; Mass of clamp 3: 0.01643 kg Mass 3: 0.51678 kg

VI. Calculations

1.

Case 1:

- Weight 1: 0.11662 kg * 9.81 m/s^2 = 1.144 N

- Torque 1: 1.144 N * 0.402 m = 0.4599 Nm (counterclockwise)

- Weight 2: 0.211862 kg * 9.81 m/s^2 = 2.0784 N

- Torque 2: 2.0784 N * 0.228 m = 0.4739 Nm (clockwise)

- Total torque: 0.4599 Nm - 0.4739 Nm = -0.014 Nm

Case 2:

- Weight 1:

- Torque 1:

- Weight 2:

- Torque 2:

- Total torque: