A 250 kg weight is positioned 6 m from a fulcrum. If a 150 kg weight is placed at the opposite end of the balance, how far from the fulcrum should it be positioned?

Question

johnweldon1993 · Answer

I'm assuming you want Torque to = 0 since you don't want it to rotate?

johnweldon1993 · Answer

If this is true...then draw your diagram....and FBD and see that the only force acting on the fulcrum *that we want* is the force of gravity acting on he 2 masses...those will be the 'F' in your Torque equation....r*F*sin(theta)

r is going to be your distance from axis of rotation

agent0smith · Answer

The torques will have to be equal, so weight*distance must be equal on each side (you can drop the gravity g, since it'll cancel off anyway)

250*6 = 150*x

agent0smith · Answer

where x is the distance the 150kg is from the fulcrum.

johnweldon1993 · Answer

sin(theta) is going to be 90degrees because the forces are acting perpendicular to the object...and yes as @agent0smith has shown....just solve that equation...and you'll be done....my way is the *long* way for this haha

agent0smith · Answer

@johnweldon1993 your way is the physics way, which is correct, but I'm assuming this question is from a geometry or algebra class (strangely, I've seen algebra students with these questions before...)

anonymous · Answer

Therefore the answer would be 10m @agent0smith

agent0smith · Answer

10m looks about right.

anonymous · Answer

Thank you @agent0smith  and @johnweldon1993

johnweldon1993 · Answer

Wow this was in geometry? haha more advanced school than I went to....Thanks @agent0smith and yes as a confirmation @kaileigh188  10m from the fulcrum is correct