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 homeow

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 d=0.233 m from the rock, which has a mass of 385 kg, and fits one end of the rod under the rock’s center of weight.

If the homeowner can apply a maximum force of 679 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.

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Vodka 5 years 2021-07-30T03:21:26+00:00 1 Answers 67 views 0

Answers ( )

  1. Sapo1
    0
    2021-07-30T03:22:31+00:00
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    Answer:

    1.52 m

    Explanation:

    We are given that

    Maximum force=F=679 N

    Mass of rock ,m=385 kg

    Distance,d=0.233 m

    We have to find the  minimum total length L of the rod required to move the rock.

    Torque on rock=T_1=mgd=385\times 9.8\times 0.233=879.11 N

    Where g=9.8m/s^2

    Torque on man=T_2=Force\times distance=(L-d)\times 679=(L-0.233)\times 679

    T_1=T_2

    (L-0.233)\times 679=879.11

    L-0.233=\frac{879.11}{679}=1.29

    L=1.29+0.233=1.523\approx 1.52 m

    Hence, the minimum total  length of rod=1.52 m

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