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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=0.233 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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lublana

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=[tex]T_1=mgd=385\times 9.8\times 0.233=879.11 N[/tex]

Where [tex]g=9.8m/s^2[/tex]

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

[tex]T_1=T_2[/tex]

[tex](L-0.233)\times 679=879.11[/tex]

[tex]L-0.233=\frac{879.11}{679}=1.29[/tex]

[tex]L=1.29+0.233=1.523\approx 1.52 m[/tex]

Hence, the minimum total  length of rod=1.52 m

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