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 = 0.233 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 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.
Equate the two torques either side of the fulcrum:
525 * 9.8 * 0.233 = 679 * x
x = 1.76 m
then the answer is = 1.76 + 0.233 = 1.993 m
A homeowner is trying to move a stubborn rock from his yard which has a mass...
Resources 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 0.266 m from the rock, which has a mass of 325 kg, and fits one end of the rod under the rock's center of weight If the homeowner can apply a maximum force...
Need help please! thanks A homeowner is trying to move a stubbom rock from his yard which has a mass of 445 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 0.277 m from the rock so that one end of the rod fits under the rock's center of weight. If the homeowner can apply...