# Mechanics - Help?

• April 29th 2010, 03:27 PM
daveypi
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?
• April 29th 2010, 09:39 PM
Rpellar
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.