1. ## Mechanics - Help?

A couple of mechanics questions I jsut cant get my head around;

1) A light rod of 50cm lies on a horizontal table. A person holds the end of the rod. The rod is subjected to a couple, of moment 40Nm, causing it to rotate upon the table. What force perpendicular to the rod must the person exert through each hand in order to prevent rotation?

2) A uniform beam AB of mass 10kg and length 4m rests horizontally on two supports, one at A and the other 1m from B. Where must a body of mass 50kg be placed on the beam if it wished to make the reactions at the supports equal?

2. 2) An upward reaction occurs at both supports, Fa and Fb.

A downward weight force of $Fw = 10(9.81) = 98.1 N$ occurs at the center of the beam.

Assume the body of mass is placed to the left of the beams center.
It has a downward force of $Fm = 50(9.81) = 490.5 N$

The sum of the forces in the Y (vertical) direction is 0.
Therefore we have, $Fa + Fb = 490.5 N + 98.1 N = 588.6 N$
For $Fa = Fb$, we have 2 $Fa = 588.6N$
Therefore $Fa = Fb = 294.3 N$

The sum of the moments about the beams center is 0 (CCW is positive).
Therefore we have, $Fm(x) - Fa(2m) + Fb(1m) = 0$
Subbing in values for Fa, Fb, and Fm and solving for x, we get $x = 0.6 m$
Thus the mass should be placed 0.6 meters to the left of the beams center so that the reaction forces can equal eachother.