A homeowner is trying to move a rock that ways 345 kg. by using a a lever arm an

Question

A homeowner is trying to move a rock that ways 345 kg. by using a a lever arm and a fulcrum. The fulcrum is at d=0.211m from the rock so rod fits under center of rocks weight.
If homeowner applies max force of 663 N at other end of rod what is the minimum total length in meters of rod to move rock. Assume rod is massless and nearly horizontal so weight of rock and homeowners force are essentially vertical. A homeowner is trying to move a rock that ways 345 kg. by using a a lever arm and a fulcrum. The fulcrum is at d=0.211m from the rock so rod fits under center of rocks weight.
If homeowner applies max force of 663 N at other end of rod what is the minimum total length in meters of rod to move rock. Assume rod is massless and nearly horizontal so weight of rock and homeowners force are essentially vertical.
If homeowner applies max force of 663 N at other end of rod what is the minimum total length in meters of rod to move rock. Assume rod is massless and nearly horizontal so weight of rock and homeowners force are essentially vertical.

Explanation / Answer

mass of rock=345kg

W = weight of the rock = mg

=345*9.8

=-3381N (The weight points down.)

d = length of the moment arm of the rock = 0.21 m
x = length of the moment arm of the homeowner = to be determined
L = total length of the rod = d + x = 0.21 m + x = to be determined

Fh = downward force exerted by the homeowner = -663 N

F.d=-FH*x

x=3381*0.21/663

=1.07m
the minimum total length in meters of rod to move rock.

=d+x=0.21+1.07

=1.28m

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