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 = 0.299 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 711 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

Load = 365 Kg = 365*9.81 = 3580.65 N

Effort = 711 N

Load arm (distane between rock and fulcrum) = 0.299 m

Effort arm (distance between fulcrum and where homwowner applies effort) = l

we know, load * load arm = effort * effort arm

=> 3580.65 * 0.299 = 711 * l

=> l = 1.506 m

So minimum total length of rod required = load arm+effort arm = 0.299 + 1.506 = 1.805 m

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