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A mass less rigid rod of length 3d is pivoted at a fixed point W, and two forces each of magnitude F are applied vertically upward as shown. A third vertical force of magnitude F may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude F is applied at point

(A) W only 

(B) Y only 

(C) V or X only 

(D) V or Y only 

(E) V, W, or X 

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Correct option (C) V or X only 

On the left of the pivot Ʈ= Fd, on the right side of the pivot Ʈ = F(2d). So we either have to add 1(Fd) to the left side to balance out the torque or remove 1(Fd) on the right side to balance out torque. Putting an upwards force on the left side at V gives (2Fd) on the left to balance torques, or putting a downwards force on the right side at X give a total of Fd on the right also causing a balance.

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