Asked by Mike

A homeowner is trying to move a stubborn rock from his yard which has a mass of 1025 kg. By using a lever arm (a piece of wooden board) and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum 24.4 cm from the rock so that it fits under the rock\'s center of weight. If the homeowner can apply a maximum force of 696.0 N, what is the minimum total length of board required to move the rock?
Answered by bobpursley
696*A=1025g*.244

solve for A, then length= L+.244 cm
Answered by drwls
The lever arm length from fulcrum to the end of the board is L. It must satisfy the moment-balance equation

L*696 = (1025 g)*0.244

g = 9.8 m/s^2

Solve for L. Add 0.244 m to L for the length of the board.
Answered by Mike
Thank you!!
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