.
A uniform beam 3m long weights 100 N. Loads of 50 N and 150 N are placed on the
beam at points which are ½ m and 2½ m, respectively, from its left end.
If the beam is kept in a horizontal position by supports at its two ends, find these reactions.
~~~~~~~~~~~~~~
Let x be the reaction force at the left end support, and
Let y be the reaction force at the right end support.
Then we have two equations.
One equation says that the total weight is at equilibrium
x + y = 100 + 50 + 150 newtons,
or
x + y = 300. (1)
The second equation says that the total rotational moments about the central point of the beam is zero
(no rotation). The distances from the central points to the loads BOTH are 1.00 m.
The weight of the beam is distributed uniformly and therefore creates ZERO rotational moment, so we can forget about it.
THEREFORE, the equation of rotational moment takes the form
-x*1.5 + 50*1 + y*1.5 - 150*1 = 0, ( ! take into account the signs of the moments ! )
or
-1.5x + 1.5y = 150*1 - 50*1
-1.5x + 1.5y = 100
-x + y = 66 . (2)
Equations (1) and (2) are easy to solve. They imply y = 183.33 N, x = 116.67 N, which is your answer.
ANSWER. The reaction forces are 116.67 N (left end at x= 0) and 183.33 N (right end at x= 3 m).
Solved.