Torque; where to place axis (levers) question.

  • #1
This problem involves torque, and levers. I am so stuck on this, I really have no idea where to even begin. The problem states:

"Tom weighs 150lb and he wants to lift a 600lb stone lid form a well. We give him a 12ft, unbreakable lever that has one end attached to the stone (at the COM), and a secure axis. He positions the lever at 30 degrees above horizontal to the ground, and places his hands 6 inches from the end of the lever. He hangs from the lever without "jumping around". Where would he need to place the axis to enable him to lift the stone lid?


I dont even know where to start with this problem. It is in our powerpoint on torques. I tried going to get help for this problem at tutoring, but was still unable to understand how to do the problem.
 

Answers and Replies

  • #2
BiGyElLoWhAt
Gold Member
1,573
118
Ok so T=R cross F
or
RFsin(theta), right?

gravity exerts a torque of r_stone * 600lbs * sin(60degrees) and Tom exerts a torque of r_tom * 150lbs* sin(60degrees) and those are the only 2 torques.

to get started, set up a newtons second law equation (sum the torques) and if you want no rotation, set it equal to 0. doing so will say that tom is exerting the same torque on the lever as the booulder is. makes sense right? I think that should get you going, and hopefully help you figure it out.

*HINT* r_boulder = 12ft - r_tom
 

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