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 445 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 = 0.299 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 631 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 Im Ti1

Explanation / Answer

According to the concept of the rotational dynamics

The balancing reaction

F1*d=F2*L

So now we find the minimum length to rock moves

Here we observed the weight of the rock act left side and the maximum force applied right side

Mg*d=F2*L

445*9.8*0.299=631*L

Therefore the minimum length required to the rock moves

L=445*9.8*0.299/631=0.204m

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