A horizontal 2.0-kg rod is 2.0 m long. An 9.0-kg block is suspended from its lef

Question

A horizontal 2.0-kg rod is 2.0 m long. An 9.0-kg block is suspended from its left end, and a 6.0-kg block is suspended from its right end.

Part A

Determine the magnitude of the single extra force necessary to keep the rod in mechanical equilibrium.

Express your answer with the appropriate units.

Part B

Determine the direction of the single extra force necessary to keep the rod in mechanical equilibrium.

Determine the direction of the single extra force necessary to keep the rod in mechanical equilibrium.

Part C

At what distance from the left end of the rod must the point of application of this force be?

Express your answer with the appropriate units.

Explanation / Answer

Mr = 2.0 Kg
Lr = 2.0 m
Ml = 9.0 Kg
Mr = 6.0 Kg
Let the unkown mass be M ?

Vertical Forces should be in balance,
(2.0 + 9.0 + 6.0) * 9.8 = F
F = 166.6 N
Force, F = 166.6 N

Direction of this Force = Upwads !!

For rod to be in equilibrium about it's Left End,
6.0 * 2.0 *9.8 + 2.0 * 1.0 * 9.8 = 166.6 * x
x =  0.824 m
Distance from the left end, x =  0.824 m

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