Question #98539

A weightless rod CD, 1.0m long is acted upon by horizontal forces of 2N, 4N, 8N is shown in the figure. What force will be necessary to produce equilibrium?

Expert's answer

Since there is no figure shown, hope these basic principles will help to solve the real problem. 8N acts on the left edge down, and on the right 2N - up and 4N - down.


Moment of force equals force times distance (from the point where the force is applied to the axis of rotation):

M=FdM=F\cdot d

Forces that try to move a rod upward are positive, downward - negative. For equilibrium, a sum of moments on the left must be equal to the one on the right:

8Nl2+xl2=2Nl24Nl2-8N\cdot \frac{l}{2}+ x\cdot \frac{l}{2} =2N\cdot \frac{l}{2} - 4N\cdot \frac{l}{2}

From this we get that x is the force acting upward on the left edge and is equal to 6N.



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