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 pivotpoint) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance d= 0.244 mfrom the rock, which has a mass of 365 kg, and fits one end of the rod under the rock's center of weight.If the homeowner can apply a maximum force of 695 N at the other end of the rod, what is the minimum total length L of therod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock andhomeowner's force are both essentially vertical. The acceleration due to gravity is g = 9.81 m/s?.%3DL =

Given Distance of fulcrum from rod d=0.244m Mass of rock m=365kg Maximum force applied by homeowner…

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 d = 0.244 m from the rock, which has a mass of 365 kg, and fits one end of the rod under the rock's center of weight.

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 d = 0.244 m from the rock, which has a mass of 365 kg, and fits one end of the rod under the rock's center of weight.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter10: Rotational Motion

Section: Chapter Questions

Problem 79P

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Similar questions

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