Equilibrium

In a nutshell

A body which is in equilibrium has a resultant force of $$0N$$​ and resultant moment of $$0Nm$$​. 

Equations 

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Variable definitions

Finding equilibrium

When solving problems with rigid bodies, you can choose to take moments about a point. When this body is in equilibrium, the resultant moment at the point is zero. This means that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.

Example 1

A uniform rod weighing $$20N$$ rests on supports $$A$$ and $$B$$.  Calculate the magnitude of the reaction at each of the supports.

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The rod is uniform, therefore the weight acts at the centre of mass. The rod is in equilibrium therefore the total forces acting downwards equals the total forces acting upwards. 

Take moments about point $$A$$. The centre of mass is $$1m$$ away from this point:

$$\begin{aligned}20\times 1&= N_B\times 2.5\\ 20&=2.5N_B\\ 8&=N_B\end{aligned}$$​​

The upwards forces are equal to the downwards forces, therefore: 

$$\begin{aligned}N_A+N_B&=20\\ N_A+8&=20\\ N_A&=12N\end{aligned}$$​​

Therefore: 

The reaction at point $$A$$ is $$\underline{12N}$$ and the reaction at point $$B$$ is $$\underline{8N}$$.

Example 2

The diagram shows a light rod in equilibrium. Find the values of the unknown forces.

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The rod is light, therefore its weight is negligible.

Take moments about a point 'A', at the far left end of the rod. If the rod is in equilibrium, the resultant moment at this point is $$0$$:

$$1X+3(15)=4Y$$​​

Take moments about a point 'B' at $$15N$$ to create another equation for $$X$$ and $$Y$$:

$$2X+1Y=3(20)$$​​

Set up a pair of simultaneous equations: 

$$\begin{aligned}X+45&=4Y\\ 2X+Y&=60\end{aligned}$$​​

Solve the simultaneous equations:

$$\begin{aligned}X&=4Y-45\\ 2(4Y-45)+Y&=60\\8Y-90+Y&=60\\ 9Y&=150\\ Y&=\dfrac{50}{3}=16.7\\X&=4\bigg(\dfrac{50}{3}\bigg)-45\\ X&=\dfrac{65}{3}=21.7\end{aligned}$$​​

Therefore, $$X$$ is $$\underline{21.7}$$ and $$Y$$ is $$\underline{16.7}$$. ​