A 44 -N crate is suspended from the left end of a plank. The plank weighs 21 N,

Question

A 44-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity does not lie halfway between the two ends. The system is balanced by a support that is 0.30 m from the left end of the plank. How far to the right of the support is the plank's center of gravity?

Distance of center of gravity from support = m

A 44-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity does not lie halfway between the two ends. The system is balanced by a support that is 0.30 m from the left end of the plank. How far to the right of the support is the plank's center of gravity? Distance of center of gravity from support =

Explanation / Answer

Let it be x m right from support

Force equilbrium conditions

Fs = Mg +44, where Fs is due to support

Fs = 21+44 = 65 N upward

Mg downward at centre of gravity, 44 N downward at left end

Torque equilibrium

Balancd torque about centre of gravity,

so torque due to mg (=21N) will be zero

So torrque due to 44N at centre of gravity + torrque due to Fs = 65 at centre of gravity = 0

44*(0.3+x) = 65*x

13.2 = 21x

x = 13.2/21 = 0.62857 m right from support

SO centere of gravity is at 0.62857 m right from support

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