1. A horizontal meter stick with a scale that increases from left to right has a

Question

1. A horizontal meter stick with a scale that increases from left to right has a clamp at the 0.370m mark(37.0cm) from which a hanger with a weight is hung. The clamp has a mass of 15.00g, the mass hanger is marked 50g and the weight has a mass of 38.00g. The entire system is static; it is in rotational equilibrium. Remember that, in such a situation, any point can be thought of as the pivot point with torques being calculated about it.

a) What is the length in meters of the lever arm from the 0.00m mark to the position of the clamp?

b) What torque (in N*m) does the weight at the 0.37m mark produce about the 0.00m mark? (Hint: make sure the sign is correct.)

c) What is the length in meters of the lever arm from the 0.500m mark to the position of the clamp?

d) What torque(in N*m) does this produce about the 0.500m mark?

2. Assume the center of gravity of the meter stick (the point at which the weight of the entire meter stick seems to be concentrated) is at the 0.500m mark. The meter stick is pivoted on a fulcrum at the 0.350m mark and balanced by placing a total mass of the 80 g at the 0.100m mark. What is the mass of the meter stick? Recall that according to the definition of the center of mass(gravity), the mass of the meter stick may be treated as though it is all at the center of mass. Make a sketch of the situation before making the calculation.

3. A meter stick has a mass of 98.00g and a center of gravity at position 0.492m. A mass of 130.00g is placed at position 0.927m, a mass of 50.00g at 0.277m and a mass of 83.20g at 0.120m. The meter stick is pivoted on a fulcrum at 0.600m. Find all the torques about the pivot (in N*m) with counter clockwise torques indicated as positive and clockwise torques as negative. Equilibrium is not assumed in this case so the torques need not add to zero.

Mass Position(Meters)

1

2

3

4

Pivot Position(Meters)

1

2

3

4

Lever Arm (Meters)

1

2

3

4

mass (kg)

1

2

3

4

force (N)

1

2

3

4

torque (N*m)

1

2

3

4

Explanation / Answer

1

a) 0.37 m

b) W = mg = (50+38+15)*10^-3*9.81 = 1.0104 N

Torque = -F*r = -1.0104*0.37 = -0.37386 Nm

c)

r = 0.5 - 0.37 = 0.13 m

d)

Torque = -F*r = -1.0104*0.13 = -0.1313 Nm

2.

80*(0.35 - 0.1) = m*(0.5 - 0.35)

m = 133.33 grams

3.

T1 = 98*10^-3 *9.81 *(0.6 - 0.492) = 0.104 Nm

T2 = -130*10^-3 *9.81 *(0.927 - 0.6) = -0.417 Nm

T3 = 50*10^-3 *9.81 *(0.6 - 0.277) = 0.1584 Nm

T4 = 83.2*10^-3 *9.81 *(0.6 - 0.12) = 0.3918 Nm

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