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 = 0.288 m from the rock, which has a mass of 345 kg, and fits one end of the rod under the rock's center of weight. Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical. The acceleration due to gravity is g = 9.81 m/s². If the homeowner can apply a maximum force of 615 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? L = m

Principles of Physics: A Calculus-Based Text
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Chapter10: Rotational Motion
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Problem 28P: A uniform beam resting on two pivots has a length L = 6.00 m and mass M = 90.0 kg. The pivot under...
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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 = 0.288 m
from the rock, which has a mass of 345 kg, and fits one end of the rod under the rock's center of weight.
d
Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both
essentially vertical. The acceleration due to gravity is g
= 9.81 m/s².
If the homeowner can apply a maximum force of 615 N at the
other end of the rod, what is the minimum total length L of
the rod required to move the rock?
L =
m
Transcribed Image Text: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 = 0.288 m from the rock, which has a mass of 345 kg, and fits one end of the rod under the rock's center of weight. d Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical. The acceleration due to gravity is g = 9.81 m/s². If the homeowner can apply a maximum force of 615 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? L = m
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