# IMAT 2016 Q58 [Upwards Forces]

The diagram shows a uniform horizontal beam of negligible mass, 5.0 m long, placed on two supports, one at each end. It has a 300 N weight placed 1.0 m from one end and a 500 N weight placed 1.0 m from the other end. Both weights act vertically on the beam as shown in the diagram.

What are the upward forces from the two supports acting on the beam?

A. 340 N and 460 N
B. 300 N and 500 N
C. 240 N and 560 N
D. 400 N and 400 N
E. 360 N and 540 N

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We can start off by remembering a simple rule in moments. When a bar is in equilibrium- we know that the total upwards force must equate the total downwards force.

This means that in this case, our total upwards and downwards force must total 800N- stemming from 300N + 500N. We can easily eliminate E at this point.

We can express this idea by saying “x+y= 800N”. Where “x is 300N” and “y is 500N”.

The next step is to visualize each moment separately and construct an equation or each side. For this we can look at this diagram:

Here we take the moment of Y:

We can therefore say that 5y= 500+ (300x4).
Where 5y is the moment of the total beam, 5m multiplied by a Force y.

Quickly solving this equation we can formulate that 5y=1700 and therefore y= 340N.

We can replicate this exact solving for the opposite side. Let’s look at this visual:

Here we take the moment of X:

We can therefore say that 5x= 300+ (500x4).
Where 5y is the moment of the total beam, 5m multiplied by a Force y.

Quickly solving this equation we can formulate that 5x= 2300 and therefore x= 460N.