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 d=0.233 m from the rock, which has a mass of 385 kg, and fits one end of the rod 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.

Answer 1

Answer:

1.52 m

Explanation:

We are given that

Maximum force=F=679 N

Mass of rock ,m=385 kg

Distance,d=0.233 m

We have to find the  minimum total length L of the rod required to move the rock.

Torque on rock=$T_1=mgd=385\times 9.8\times 0.233=879.11 N$

Where $g=9.8m/s^2$

Torque on man=$T_2=Force\times distance=(L-d)\times 679=(L-0.233)\times 679$

$T_1=T_2$

$(L-0.233)\times 679=879.11$

$L-0.233=\frac{879.11}{679}=1.29$

$L=1.29+0.233=1.523\approx 1.52 m$

Hence, the minimum total  length of rod=1.52 m