Ahomeowner 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.

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

Torque of man

Torque of rock

The torques acting on the system is conserved

The length of the rod is 1.17894 m

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft