1. The center of gravity of a meter stick that weighs 1.950 N is at the 50.00 cm

Question

1. The center of gravity of a meter stick that weighs 1.950 N is at the 50.00 cm mark. A 3.54 N weight is suspended from the 20.0 cm mark and an 8.00 N weight is suspended from the 70.0 cm mark on the meter stick. A rough diagram is shown Figure 7-37. Where would you place the fulcrum to balance this system? (Calculate the exact location of the center of gravity (x) of the system.)

2.What is the net torque on a pulley if there is a 25.0 N counterclockwise force at 4.01 cm from the pivot and a 4.1 N friction force on the axle 1.80 cm from the pivot point. Give the answer in Nm and work to 2 decimal places.

Explanation / Answer

Let X be the distance for the fulcrum.

The moment acting of the left side of the fulcrum = 8 N * (70 cm - X)
The moments acting on the right side of the fulcrum = 1.95 N * (X - 50 cm) + 3.54 N * (X - 20 cm)
Since at equilibrium the moments have to be equal
8 N * (70 cm - X) = 1.95 N * (X - 50 cm) + 3.54 N * (X - 20 cm)

X=16.62cm

2. F1=25N,r1=4.01cm

F2=4.1N.r2=1.8cm

Fnet=r1xF1+r2xF2=25*4.01+4.1*1.8=107.63N-cm=1.08N-m

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