Question



The 4.6 kg, uniform, horizontal rod in (Figure 1) is seen from the side. Part A What is the gravitational torque about the po
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Answer #1

Calculate the Weight of Rod = mg
w = 4.6 * 9.81
w = 45.13 N

Calculate the Centre of gravity of Left side :
CGl = 25/2 = 12.5 from pivot

Calculate the Weight of left section :
Wl = 45.13*25/100
Wl = 11.28 N

Write the equation for torque on left side.

Torque = Force*Distance
Tl = 11.28*12.5
Tl = 141 N.cm will be anticlockwise

Calculate the Centre of gravity of Right side :
CGr = 75/2 = 37.5 from pivot

Calculate the Weight of right section :
Wr = 45.13 * 75/100
Wr = 33.85 N

Write the equation for torque on left side.

Torque = Force*Distance
Tr = 33.85 * 37.5
Tr = 1269.28 N.cm will be clockwise

Tnet = Tr - Tl
= 1269.28- 141
        = 1128.38 N.cm
   = 11.28 N.m

The positive sign of the net torque indicates the clockwise direction.

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