A homeowner is trying to move a stubborn rock from his yard which has a mass of

Question

A homeowner is trying to move a stubborn rock from his yard which has a mass of 345 kg, By using a lever arm (a piece of metal rod) and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum d= 0.244 m from the rock so that one end of the rod fits under the rock's center of weight. If the homeowner can apply a maximum force of 623 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical Number

Explanation / Answer

Using torque balance about center of weight

Tnet = Ts - Th = 0

Ts = Th

Torque is given by

T = F*r

F = m*g

Ms*g*d = Fh*r

where r = length of rod from Center of weight to homeowner

Fh = maximum applied force by homeowner

Ms = mass of stone

r = Ms*g*d/Fh

Using given values:

r = 345*9.81*0.244/623

r = 1.325 m

Total length of rod will be

L = d + r = 0.244 + 1.325 = 1.568 m

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