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Equilibrium

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Summary

Equilibrium

In a nutshell

A body which is in equilibrium has a resultant force of 0N0N0N​ and resultant moment of 0Nm0Nm0Nm​. 


Equations 

DESCRIPTION
EQUATION
Moment of FFF​ about a pivot, where FFF and ddd are perpendicular.​
​M=F×dM = F \times dM=F×d​​
Moment of FFF about a pivot, where FFF and ddd are not perpendicular.
​M=F×dsin⁡θM = F \times d \sin\thetaM=F×dsinθ​​

​

Variable definitions

QUANTITY NAME
SYMBOL
UNIT NAME
UNIT
​MomentMomentMoment​​
​MMM​​
​Newton metreNewton \space metreNewton metre​​
​NmNmNm​​
​ForceForceForce​​
​FFF​​
​NewtonNewtonNewton​​
​NNN​​
​DistanceDistanceDistance​​
​ddd​​
​MetreMetreMetre​​
​mmm​​



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 20N20N20N rests on supports AAA and BBB.  Calculate the magnitude of the reaction at each of the supports.

​

Maths; Moments; KS5 Year 13; Equilibrium


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 AAA. The centre of mass is 1m1m1m away from this point:

20×1=NB×2.520=2.5NB8=NB\begin{aligned}20\times 1&= N_B\times 2.5\\ 20&=2.5N_B\\ 8&=N_B\end{aligned}20×1208​=NB​×2.5=2.5NB​=NB​​​​


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

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


Therefore: 


The reaction at point AAA is 12N‾\underline{12N}12N​ and the reaction at point BBB is 8N‾\underline{8N}8N​.


Example 2

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


Maths; Moments; KS5 Year 13; Equilibrium

​

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 000:


1X+3(15)=4Y1X+3(15)=4Y1X+3(15)=4Y​​


Take moments about a point 'B' at 15N15N15N to create another equation for XXX and YYY:


2X+1Y=3(20)2X+1Y=3(20)2X+1Y=3(20)​​


Set up a pair of simultaneous equations: 


X+45=4Y2X+Y=60\begin{aligned}X+45&=4Y\\ 2X+Y&=60\end{aligned}X+452X+Y​=4Y=60​​​


Solve the simultaneous equations:


X=4Y−452(4Y−45)+Y=608Y−90+Y=609Y=150Y=503=16.7X=4(503)−45X=653=21.7\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}X2(4Y−45)+Y8Y−90+Y9YYXX​=4Y−45=60=60=150=350​=16.7=4(350​)−45=365​=21.7​​​



Therefore, XXX is 21.7‾\underline{21.7}21.7​ and YYY is 16.7‾\underline{16.7}16.7​. ​




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FAQs - Frequently Asked Questions

What does a light rod mean?

If a rod is light, its weight is negligible.

What does 0 resultant moment mean?

This means that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.

What does it mean when a body is in equilibrium?

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

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