A uniform, 255-N rod that is 2.00 m long carries a 225-N weight at its right end

Question

A uniform, 255-N rod that is 2.00 m long carries a 225-N weight at its right end and an unknown weight W toward the left end . When W is placed 30.0 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75.0 cm from the right end.

(a) Find W and explain why you used the method you did to find it.

(b) If W is now moved 25.0 cm to the right, how far and in what direction must the fulcrum be moved to restore balance and why?

A uniform, 255-N rod that is 2.00 m long carries a 225-N weight at its right end and an unknown weight W toward the left end . When W is placed 30.0 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75.0 cm from the right end. Find W and explain why you used the method you did to find it. If W is now moved 25.0 cm to the right, how far and in what direction must the fulcrum be moved to restore balance and why?

Explanation / Answer

By Balancing Moments

75*225 = W*(200 - 30) + 255*25

W = 61.765 N

As For Equilibrium , the moments must balance.So I choose this method

Let it is at a Distance x from the middle towards Right

Therefore

61.765*(100-30-25 + x) + 255*x = 225*(100 - x)

x = 36.4 cm

So it is moved towards Right by Distance 11.4 cm from the previous position

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