A uniform, 237-N rod that is 1.96 m long carries a 225-N weight at its right end

Question

A uniform, 237-N rod that is 1.96 m long carries a 225-N weight at its right end and an unknown weight W toward the left end (see the figure (Figure 1)). When W is placed 50.1 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 71.0 cm from the right end. Find W If W is now moved 29.8 cm to the right, how far must the fulcrum be moved to restore balance? If W is now moved 29.8 cm to the right, in what direction must the fulcrum be moved to restore balance?

Explanation / Answer

A. The equilibrium condition is that the total momentum of all forces should be 0.

225=0.71=237(L/2-0.71)+W(L-0.501-0.71)

L=1.96 m

Solve this eq and get W: ~127.85 N.

B.

The equilibrium condition is that the total momentum of all forces should be 0.

Consider the distance (unknown) from the right extreme point to the new position of the fulcrum (distance between 225 N weight to the new position of the fulcrum) being x.

We have the eq:

Fg2x=Fg1(L/2-x)+W*(L-r1-r2-x)

r1=0.501 m

r2=0.298 m

L=1.96 m

Fg2=225N

Fg1=237 N

W=295 N

Replace in (1) and get:

x~ 0.3938 m

The fulcrum should be moved 0.71-0.3938~0.3162 m=3.162 cm

C.

Also to the right.

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