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 365 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 = 0.277 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 703 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. L=

Explanation / Answer

here,

mass of rock , m = 365 kg

d = 0.277 m

applied force , f = 703 N

let the length of rod be L

to move the rock,

m*g*d = f*( l - d)

365 * 9.8 * 0.277 = 703 * ( l - 0.277)

l = 1.7 m

the minimum total length of rod L is 1.7 m

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