3. A meter stick has a mass of 98.00g and a center of gravity at position 0.492m

Question

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

(m)

Pivot Positon

(m)

Lever arm

(m)

Mass

(kg)

Force

(N)

Torque

(N*m)

mass position

(m)

Pivot Positon

(m)

Lever arm

(m)

Mass

(kg)

Force

(N)

Torque

(N*m)

Explanation / Answer

mass position

(m)

Pivot Positon

(m)

Lever arm (r)

(m)

Mass
(m)

(kg)

Force

(N)
F = mg

Torque

(N*m)
N = r x F

Net torque = - 0.416598 + 0.3913728 +.15827 + 0.1037232 = 0.236768 N m

mass position

(m)

Pivot Positon

(m)

Lever arm (r)

(m)

Mass
(m)

(kg)

Force

(N)
F = mg

Torque

(N*m)
N = r x F

0.927 0.6 0.927 - 0.6 = 0.327 0.13 1.274 - 0.416598 0.12 0.6 0.6 - 0.12 = 0.48 0.0832 0.81536 0.3913728 0.277 0.6 0.6 - 0.277 = 0.323 0.05 0.49 0.15827 0.492 0.6 0.6 - 0.492 = 0.108 0.098 0.9604 0.1037232

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