Question

A uniform, 246-N rod that is 2.08 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 47.9 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 65.1 cm from the right end.

A) Find W

B) If W is now moved 24.7 cm to the right, how far must the fulcrum be moved to restore balance?

C) If W is now moved 24.7 cm to the right, in what direction must the fulcrum be moved to restore balance?

Answer 1

Part A.

Given that fulcrum is balanced in this scenario, So that center of gravity of rod will be at fulcrum,

Now center of gravity in horizontal direction is given by

Xcm = (W1*x1 + W2*x2 + W3*x3)/(W1 + W2 + W3)

W1 = Weight placed on the right end of rod = 225 N

W2 = Weight of rod = 246 N

W3 = Weight placed at 'x' distance from left end = W

(Here we assume that origin is at the left end of rod, then)

x1 = position of W1 w.r.t. to left end = L = 2.08 m

x2 = position of of center of mass of rod from left end = 2.08 m/2 = 1.04 m

Xcm = position of center of gravity of rod at fulcrum from left end = L - 65.1 cm = 2.08 m - 0.651 m = 1.429 m

x3 = position of W from left end = 47.9 cm = 0.479 m

Using these values:

1.429 = (225*2.08 + 246*1.04 + W*0.479)/(225 + 246 + W)

Solving above equation:

W = (225*2.08 + 246*1.04 - 471*1.429)/(1.429 - 0.479)

W = 53.45 N = unknown Weight

Part B.

Now when fulcrum is moved 24.7 cm to the right then

x3 = position of W from left end = 47.9 cm + 24.7 cm = 72.6 cm = 0.726 m

Values of W1, W2, W, x1 and x2 will not change, So using these value new center of gravity will be:

Xcm = (W1*x1 + W2*x2 + W3*x3)/(W1 + W2 + W3)

Xcm = (225*2.08 + 246*1.04 + 53.45*0.726)/(225 + 246 + 53.45)

Xcm = 1.454 m

So fulcrum should be moved (1.454 - 1.429) = 0.025 m = 2.5 cm

Since we have moved W in right direction, So fulcrum should also be moved in right direction