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 525 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 = 0.266 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 671 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. & 412412017 06:00 PM 50/100 4/24/2017 o4:06 PM Print Calculator Periodic Table 6 Gradebook Question 8 of 10 Sapling Learning Mapoeit A homeowner is trying to move a stubborn rock from his yard which has a mass of 525 kg. By using a lever arm 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 0.266 m from the rock so that one end of the rod fits under the rock s center of weight. If the homeowner can a maximum force of 671 N at the other end of the rod, what is the minimum total length a to move the rock? Assume that the rod is massless and nearly horizontal so that the weight af the rock and force are both

Explanation / Answer

for equillibrium condition

Net torque about an axis passing through fulcrum is zero

torque = rxF

m = 525 kg, d = 0.266 m , F = 671 N

Fr*d = Fh*x

Fr = mg

525*9.8*0.266 = 671*x

x = 2.04 m

L = d+x = 0.266 +2.04

minimum total length of rod L = 2.306 m

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