Question

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 ????=0.211 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 of 695 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.

Answer 1

Answer:

1.17894 m

Explanation:

The rock is at one end of the rod which is 0.211 m from the fulcrum

F = Force

d = Distance

L = Length of rod

M = Mass of rock = 325 kg

g = Acceleration due to gravity = 9.81 m/s²

Torque

$\tau=F\times d$

Torque of man

$\tau_m=F(L-d)\\\Rightarrow \tau_m=695(L-0.211)$

Torque of rock

$\tau_r=Mg\times d\\\Rightarrow \tau=325\times 9.81\times 0.211\\\Rightarrow \tau=672.72075\ Nm$

The torques acting on the system is conserved

$\tau_m=\tau_r\\\Rightarrow 695(L-0.211)=672.72075\\\Rightarrow L-0.211=\frac{672.72075}{695}\\\Rightarrow L-0.211=0.96794\\\Rightarrow L=0.96794+0.211\\\Rightarrow L=1.17894\ m$

The length of the rod is 1.17894 m