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 345 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 = 0.244 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 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.

Explanation / Answer

m = mass of the rock = 345 kg
g = acceleration due to gravity = 9.81 m/s2
W = weight of the rock = mg = (345 kg)(9.81 m/s2) = -3,384 N (The weight points down.)
d = length of the moment arm of the rock = 0.244 m
x = length of the moment arm of the homeowner = to be determined
L = total length of the rod = d + x = 0.244 m + x = to be determined
Frock = upward force on the rock -W = -(-3,384 N) = 3,384 N
Fhomeowner = downward force exerted by the homeowner = - 663 N

|Frock ·d| = |Fhomeowner·x|

Frock·d = - Fhomeowner·x

(3,384 N)(0.244 m) = -(-663 N)(x)

(3,384 N)(0.244 m) = (663 N)(x)

[(3,384 N)(0.244 m)]/(663 N) = x

x = 1.25 m

L = d + x = 0.244 m + 1.25 m = 1.489 m

Minimum Total Length of the Rod = 1.489m or 1.5 m

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