Print Question 3 of 11 Mapdo Sapling Learning A homeowner is trying to move a st

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Print Question 3 of 11 Mapdo Sapling Learning A homeowner is trying to move a stubborn rock from his yard which has a mass of 425 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 631 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 Number im O Previous O Check Answer 0 Next 1 Exit Hint

Explanation / Answer

The maximum force of 631 N is being applied at a distance of , say , x from the fulcrum. Then, to move the rock, the torque applied by the homeowner should be more or atleast equal to the toque of the rock's weight. So, equating the torques, 631*x= 425*9.81*0.266 { where  m*g= 425*9.81 is weight of rock }

Or, x= 1.75m

So, minimum length of the rod is L= 0.266+1.75= 2.03 m

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