Question

A homeowner is trying to move a stubbom 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 0.277 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 foroe of 647 N at the other end ot the rod, what is the minimum lenqth L of the rod required to move the rock? Assume that the rod is massless and neardy horizontal so that the weight of the rock and homeowner's force are both essentially vertical Number Pivicus @ Hin

Need help please! thanks

Answer 1

Suppose the rod's length be L

balance the torques about the fulcrum

Torque, T = distance x force

Torque of man, T₁ = (L - d) x 647

Torque of rock T₂ = d x Mg = 0.277 x 445 x 9.81 = 1209.23 Nm

now both the torques should balance each other as one is clockwise and other anti clockwise

so magnitude is equal T₁ = T₂

(L - d) x 647 = 1209.23

(L - 0.277) x 647 = 1209.23

So, the length should be 2.15 m

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