The 2.4 kg , uniform, horizontal rod in (Figure 1) is seen from the side. What...

Question

Question

The 2.4 kg , uniform, horizontal rod in (Figure 1) is seen from the side.

What is the gravitational torque about the point shown? The 2.4 kg , uniform, horizontal rod in (Figure 1)

Answer 1

Answer #1

here,

Weight of Rod = mg
w = 2.4 * 9.8
w = 23.52 N

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

Weight of left section :
Wl = 23.52*25/100
Wl = 5.88 N

Torque = Force*Distance
Tl = 5.88*12.5
Tl = 73.5 N.cm will be anticlockwise

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

Weight of right section :
Wr = 23.52 * 75/100
Wr = 17.64 N

Torque = Force*Distance
Tr = 17.64 * 37.5
Tr = 551.25 N.cm will be clockwise

Tnet = Tr - Tl
Tnet = 661.5 - 73.5
Tnet= 588 N.cm
TNet = 5.88 N.m

Answer 2

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