Question

A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance ????=0.266 md=0.266 m from the rock, which has a mass of 325 kg,325 kg, and fits one end of the rod under the rock's center of weight. If the homeowner can apply a maximum force of 663 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.

Answer 1

Answer:

L = 1.545 m

Explanation:

Let the total length of the rod is L

now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum

so we will have

$mg \times d = F(L - d)$

so we have

$325\times 9.8 \times 0.266 = F(L - 0.266)$

F = 663 N

$848 = 663(L - 0.266)$

$L = 1.545 m$