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 385 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 = 0.288 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 679 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.

Explanation / Answer

mass of house owner = 385 kg

distance from the rock = 0.288 m

distance of the arm for owner = x m (say)

force applied f = 679 N

torque of rock = weight * distance = 385*9.8*0.288 = 1086.624 Nm

torque of owner = 679 x

equate both

679x = 1086.624

x = 1.6 m

total length of the rod L =d+x = 0.288+1.6 = 1.88 m

so minimum length of the rod is 1.88 m

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