Finding reactions on an L-bracket with moment and force

[Satoy]

How do you change a moment about a point into a force is my main problem. I have a force of 300 N at the top of a L shaped bracket h=240 mm. So at the other end of the L at l= 180 mm the force would be -400 N right? Because at the orgin the moment would be 72000 N*mm. So what is the reaction at the orgin and the -400 N force.

I think the reaction at the -400 N force would be 400 N but can't figure the reaction at the orgin. Please Help

Sorry for the other thread in classic

 

[OlderDan]

How do you change a moment about a point into a force is my main problem. I have a force of 300 N at the top of a L shaped bracket h=240 mm. So at the other end of the L at l= 180 mm the force would be -400 N right? Because at the orgin the moment would be 72000 N*mm. So what is the reaction at the orgin and the -400 N force.

I think the reaction at the -400 N force would be 400 N but can't figure the reaction at the orgin. Please Help

Sorry for the other thread in classic

If I have deciphered your problem correctly, the corner of the bracket is being held in place by some force. Apparently the force at the top is applied horizontally to the left, and the force of -400N is downward. The sum of the forces acting must equal zero, so the reaction at the origin must have two components to offset the other two applied forces.

 

[Satoy]

I don't know if you are thinking about it right. There is only one force to begin with the 300 N at the top going horizontal in the positive x direction. At the orgin there is a stationary pivot point and a wheel at the other end on the x axis. The -400 N force I translated to the wheel from the original 300 N force.

 

[Satoy]

I'm looking for two answers on at the wheel and the other at the orgin.

 

[OlderDan]

I'm looking for two answers on at the wheel and the other at the orgin.

OK You are treating the bracket as a lever. The -400N is a force the bracket is applying to the wheel at that one end. By Newton's third law, the wheel is applying an upward force of 400N at the end of the bracket. So far, you have two forces acting on the bracket: 300N horizontal to the right and 400N upward applied at the ends of the bracket resulting in no net moment about the pivot point. The pivot point must supply a force on the bracket that will oppose the sum of the first two forces to maintain equlibrium. Break the net force from the pivot into horizontal and vertical components and require the sum of the vertical forces and the sum of the horizontal forces acting on the bracket to be zero.