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.

[tex]\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[/tex]

The length of the rod is 1.17894 m

hamzaahmeds hamzaahmeds · Answer 2 · 2021-11-01T13:33:07+01:00

This question involves the concepts of effort, load, moment, load arm, and effort arm.

The minimum length of the rod is found to be "1.179 m".

Ideally, the output of the lever machine must be equal to the input. In other words, the moment applied by the load on the lever must be equal to the moment applied by the effort on the lever.

[tex]Moment\ of\ Effort=Moment\ of\ Load\\(Effort)(Effort\ Arm) = (Load)(Load\ Arm)\\[/tex]

where,

Effort = 695 N

Load = mg = (325 kg)(9.81 m/s²) = 3188.25 N

Load Arm = 0.211 m

Effort Arm = ?

Therefore,

[tex]Effort\ Arm(695\ N)=(0.211\ m)(3188.25\ N)\\Effort\ Arm = 0.968\ m[/tex]

Now, the minimum length of the rod (l) can be found as follows:

[tex]l = Effort\ Arm+Load\ Arm\\l = 0.968\ m+0.211\ m\\[/tex]

L = 1.179 m

Learn more about the moment here:

https://brainly.com/question/6278006?referrer=searchResults

The attached picture shows the principles of moment.