# Physics Tutorial: Moment of Force. Conditions of Equilibrium

In this Physics tutorial, you will learn:

• What is moment of force?
• What are the factors affecting the moment of force?
• What is the equation and unit of moment of force?
• What happens if the acting force is not normal to the bar?
• What happens if more than two forces are acting on a turning system?
• What happens if there is no equilibrium in a turning system?
• What if the bar is heavy and as such, we cannot neglect its weight?
• What if the bar is hanged on a string?
• Which are the two conditions of equilibrium?

## Introduction

Suppose you are trying to make a long bar stay in equilibrium on a pivot. Can you put the pivot under whatever point of the bar to set up the equilibrium? Why?

Can you hang a bar at every point and still pretend to make is stay horizontal? Why?

The above questions will get answer in this tutorial, through theoretical explanation and illustrated by various examples.

## What is Moment of Force?

Let's consider again the examples mentioned in the introduction. If we have a regular and homogenous bar and we want to make it stay horizontally in equilibrium when putting it on a pivot, the only position which grants such an equilibrium (although very frail) is at the centre of mass of the bar, i.e. at its middle, as shown in the figure. Thus, if the bar length is L, the pivot must be placed under the point C (centre of mass) which is at L/2 to set up the equilibrium. Otherwise, the bar will lean in the direction of the heaviest piece. Look at the figure: The same thing occurs when we hang a bar on a string. Such a lean is not linear; the system tends to rotate around the pivot or the hanging point instead. Therefore, a turning effect towards the support is produced. In scientific terms, this turning effect is known as "Moment of Force", in short M and it is a vector quantity.

There are two possible turning directions in such systems: clockwise and anticlockwise as shown in the figure below. When the bar is very light so that its weight is negligible (such as a polystyrene bar or similar), we can use two loads placed on both sides of the pivot to set up an equilibrium similar to the abovementioned ones.

If the loads are identical, we must place the pivot at middle of the bar, otherwise the system will lean towards the farthest load from the pivot (towards the longer arm). In such cases, we can balance the system by placing a heavier load on the other end as shown in the figure below. This means there are two factors affecting the turning effect (moment) of a force. They are:

### 1. The magnitude of force

Greater the force, easier the rotation around the turning point, i.e. greater the moment of force.

### 2. The distance from the turning point

Greater the distance from the turning point, easier the rotation around the turning point, i.e. greater the moment of force.

Combining the above factors in a single equation, we obtain for the Equation of moment of force:

### Equation 1

M = F × ∆x

where F is the perpendicular force to the bar, which acts on it and ∆x is the linear distance from the application point of the force to the turning point (pivot) of the bar.

The unit of moment of force is [N × m]. Since it is obtained from the cross product of two vectors, it cannot be equal to the unit of work (and energy), i.e. Joule. In the tutorial "Work and Energy", we have stated that Work is obtained by the dot (scalar) product of two vectors: Force and Displacement. The result therefore was a scalar. On the other hand, here we have a cross product of two vectors, which gives as a result a new vector (see the tutorial "Cross (Vector) Product of Two Vectors". Therefore, energy and moment of force cannot have the same unit; the unit of moment of force is simply Newton × metre, not Joule. In the above figure, the load on the left tend to rotate the system anticlockwise as it exerts a downward force in that part of the bar due to its weight. On the other hand, the load on the right tend to rotate the system clockwise for the same reason. When the equilibrium is settled, these two opposite moments are balanced.

Therefore, based on all said above, we infer the following conclusion regarding the equilibrium in such systems:

"A system which tends to rotate around a fixed point is in equilibrium only when its clockwise moment of force is equal to the anticlockwise one"

This means the clockwise turning effect of one force is balanced by the anticlockwise turning effect of the other force.

Mathematically, we can write:

### Equation 2

Mclockwise + Manticlockwise = 0

Or

Mclockwise = - Manticlockwise

If we substitute the corresponding quantities, we obtain

### Equation 3

F1⊥ × ∆x1 + F2⊥ × ∆x2 = 0

Or

F1⊥ × ∆x1 = -F2⊥ × ∆x2

### Example 1

A 5 kg object is placed on the left end of a 4 m massless bar as shown in the figure. The system is balanced by a 75 N force acting vertically on the right end of the bar. What is the distance of the pivot from the left end of the bar? For convenience, take g = 10 N/kg.

### Solution 1

The 5 kg object exerts a downward force equal to its weight to the left-end of the bar. Let's denote this force as F1. Thus, we have:

F1⊥ = m × g
= 5 kg × 10 N/kg
= 50 N

Let's denote the required distance by d (or Δx, no matter). Thus, the other distance, i.e. from the pivot to the right end of the bar will be 4 - d. Therefore, since we know that the object tends to rotate the system anticlockwise while the force F clockwise, it is clear that the corresponding moments of force must be equal and opposite. If we take the anticlockwise as positive, we have:

F1⊥ × ∆x1 + F2⊥ × ∆x2 = 0
F1⊥ × ∆x1 = -F2⊥ × ∆x2

If we focus only in the numerical values, not in the direction, we obtain

50 × d = 75 × (4-d)
50 × d = 300-75 × d
75d + 50d = 300
125d = 300
d = 300/125 = 2.4 m

Therefore, the pivot must be at 2.4 m away from the left end (and 4 m - 2.4 m = 1.6 m away from the right end) of the bar to have equilibrium in this system.

Remark! Do not be confused by the fact that we used a scalar approach to solve this problem. Even if we used a vector approach, the result would be the same. The only difficulty in this regard is determining the directions correctly. Thus, if we consider the usual directions, we take Δx1 as positive because it starts from the object and ends to the pivot (therefore, it extends due right), while Δx2 is taken as negative because it starts from the right end of the bar (where the force acts) and lies towards left, up to the pivot. On the other hand, we can take the two acting forces as negative as they both act downwards.

Therefore, we have

F1⊥ × ∆x1 = -F2⊥ × ∆x2
-50 × d = -(-75) × [4-(-d)]
-50 × d = 300 + 75 × d
125 × d = -300
d = -300/125 = -1.6 m

The sign minus means that the object is on the left of the pivot, as here we take the pivot as a reference point.

## What happens if the acting force is not normal to the bar?

In this case, we consider only the force component that lies normal to the bar as only it contributes in the rotation of the system. The other component, i.e. the component of force according the direction of the bar, goes in vain, as it gives no contribution in the rotation. Look at the example below:

### Example 2

How far from the bar is the application point of the force F in the system shown in the figure below? Take g = 10 N/kg, cos 530 = 0.6 and sin 530 = 0.8. ### Solution 2

Let's use a scalar approach to solve this problem. We have the following clues:

m = 4 kg
Δx1 = d1 = 1.2 m
F2 = 60 N
d2 = ?

Let's find some missing quantities first. Thus,

F1 = W = m × g
= 5 kg × 10 N/kg
= 50 N
F2x = F2 × cos 530
= 60 N × 0.6
= 36 N
F2y = F2⊥ = F2 × sin 530
= 60 N × 0.8
= 48 N

Thus, we have

F1⊥ × d1 = F2⊥ × d2

After substitution, we obtain

40 × 1.2 = 48 × d2
48 = 48 × d2
d2 = 48/48 = 1m

## What happens if more than two forces are acting on a turning system?

If there are more than two forces acting on such a system, first we determine the direction of rotation of each force. This is because they may be on the same side of the bar but act in opposite directions as shown in the figure. From the figure, you can see that F1 causes an anticlockwise turning effect in the system while F2 causes a clockwise effect despite both forces act due right of the turning point. On the other hand, F3 causes a clockwise effect. Therefore, if the system is in equilibrium, we must write (in scalar mode):

m1 = m2 + m3

m1 + m2 = m3

Therefore, we must solely consider the direction of rotation and not the direction in which a force acts in respect to the turning point.

### Example 3

Three forces are acting at the system shown in the figure below. ### Solution 3

F1 turns the system anticlockwise while F2 and F3 clockwise. Also, d2 = 60 cm = 0.6 m and d3 = 80 cm = 0.8 m.

If there is equilibrium, we have:

M1 = M2 + M3

Or

F1⊥ × d1 = F2⊥ × d2 + F3⊥ × d3

All forces are normal to the bar, so, no change is needed on them. Substituting the known values, we obtain

30 × 1 = 40 × 0.6 + F3 × 0.8
30 = 26 + F3 × 0.8
30-26 = F3 × 0.8
4 = F3 × 0.8
F3 = 4/0.8
F3 = 5N

## What happens if there is no equilibrium in a turning system?

In such a case, the system is not balanced. This means there will be a non-zero resultant moment in the direction of the greatest turning effect. Consider the example below.

### Example 4

Three forces act on the system shown below. What is the resultant moment of force if cos 300 = 0.86, sin 300 = 0.5, cos 530 = 0.6 and sin 530 = 0.8?

### Solution 4

In this system, the 100 N and the 60 N force tend to rotate the system clockwise while the 120 N force create an anticlockwise turning effect. Thus, if we index the forces (and distances from the turning point) as 1, 2, and 3 from left to right, we have for the resultant moment of force MR:

MR = M1 + M2 - M3
MR = F1⊥ × d1 + F2⊥ × d2 - F3⊥ × d3
MR = F1 × sin 300 × d1 + F2 × sin 530 × d2 - F3 × sin 900 × d3

Substituting the values, and giving that sin 900 = 1, we obtain:

MR = 100 × 0.5 × 2 + 60 × 0.8 × 3 - 120 × 1 × 6
= 100 + 144 - 720
= -476 N × m

Since we have chosen clockwise as positive, the result obtained means there is a non-zero resultant moment of force in the anticlockwise direction. Therefore, the whole system rotates anticlockwise by 476 N × m.

## What if the bar is heavy and as such, we cannot neglect its weight?

In this case, we take the weight of the bar as an additional downward force exerted at the bar's centre of gravity. Then, we use the usual approach mentioned above. Look at the following example:

### Example 5

A 4 kg uniform and homogenous bar is placed on a pivot as shown in the figure. What is the mass of the extra object we must place on the right end of the bar to set the equilibrium?

### Solution 5

The bar exerts its weight at its centre because it is uniform and homogenous (as discussed in the tutorial "Centre of Mass. Types of Equilibrium"). Therefore, we must write a force equal at

Wbar = mbar × g
= 4 kg × 9.81 N/kg

at middle of the bar, i.e. at

dbar = Ltotal/2
= 2m/2
= 1m

away from its left end. This means the centre of mass is 1 m - 0.6 m = 0.4 m on the right of the pivot, as shown below. Therefore, we have dbar = 0.4 m and dobject = 60 cm = 0.6 m.

It is obvious that the bar's weight produces a anticlockwise moment while the extra object a clockwise moment because when there is equilibrium, the moments of force are equal and opposite.

Therefore, we have:

mbar × g × dbar = mobject × g × dobject

Simplifying g from both sides, we obtain for the mass of the extra object:

mbar × dbar = mobject × dobject
mobject = mbar × dbar/dobject
= 4 × 0.4/0.6
= 2.67 kg

## What if the bar is hanged on a string?

The approach and the physical principles involved are absolutely the same as those discussed above. Consider the following example:

### Example 6

What is the mass of the object 3 (m3) in the system shown below if m1 = 6 kg? All bars and ropes are massless. ### Solution 6

From the figure, we can see that the system is in equilibrium. Therefore, we can write for the left bar:

m1 × g × 5 units = m2 × g × 1 unit

Simplifying g from both sides, we obtain after the substitutions

6 × 5 = m2 × 1
m2 = 30 kg

Therefore, we have for the two objects hanged on the small bar in the left,

mleft = m1 + m2
= 6 kg + 30 kg
= 36 kg

Now we can use the same method to work out the mass m3. Thus,

mleft × g × 2 units = m3 × g × 5 units
36 × 2 = m3 × 5
m3 = 36 × 2/5
= 14.4 kg

## Conditions of Equilibrium

In the tutorial "Newton's Second Law of Motion", we saw that an object is in equilibrium if it tends to move linearly and the forces acting on it are balanced. Thus, if we want to set in equilibrium such an object or system, we must meet the following condition:

FR = F1 + F2 + F3 + … = 0

This is known as the first condition of equilibrium. It is valid for translational (linear) motion

On the other hand, if the system tends to rotate around a fixed point, the equilibrium is settled only when the resultant moment of force is zero, i.e. when

MR = M1 + M2 + M3 + … = 0

Or

MR = F1 × ∆x1 + F2 × ∆x2 + F3 × ∆x3 + … = 0

The above equation is known as the second condition of equilibrium. It is valid for circular motion around a fixed point.

## Summary

When a system tends to rotate around the pivot or the hanging point, a turning effect towards the support is produced. In scientific terms, this turning effect is known as "Moment of Force", in short M and it is a vector quantity.

There are two possible turning directions in such systems: clockwise and anticlockwise.

There are two factors affecting the turning effect (moment) of a force. They are:

### 1. The magnitude of force

Greater the force, easier the rotation around the turning point, i.e. greater the moment of force.

### 2. The distance from the turning point

Greater the distance from the turning point, easier the rotation around the turning point, i.e. greater the moment of force.

Combining the above factors in a single equation, we obtain for the Equation of moment of force:

M = F × ∆x

where F is the perpendicular force to the bar, which acts on it and ∆x is the linear distance from the application point of the force to the turning point (pivot) of the bar.

"A system which tends to rotate around a fixed point is in equilibrium only when its clockwise moment of force is equal to the anticlockwise one."

This means the clockwise turning effect of one force is balanced by the anticlockwise turning effect of the other force.

Mathematically, we can write:

Mclockwise + Manticlockwise = 0

Or,

F1⊥ × ∆x1 + F2⊥ × ∆x2 = 0

If any of the acting force is not normal to the bar, we consider only the force component that lies normal to the bar as only it contributes in the rotation of the system. The other component, i.e. the component of force according the direction of the bar, goes in vain, as it gives no contribution in the rotation.

If there are more than two forces acting on such a system, first we determine the direction of rotation of each force. This is because they may be on the same side of the bar but act in opposite directions. Then we use the equation of moments of force.

If the system is not in equilibrium, it is not balanced. This means there will be a non-zero resultant moment in the direction of the greatest turning effect.

If the bar is too heavy to be neglected, we take its weight as an additional downward force exerted at the bar's centre of gravity. Then, we use the usual approach mentioned above to find any missing quantity. This is also true for systems in which a bar is hanged on a string and some objects are attached to it.

An object is in equilibrium if it tends to move linearly and the forces acting on it are balanced. Thus, if we want to set in equilibrium such an object or system, we must meet the following condition:

FR = F1 + F2 + F3 + … = 0

This is known as the first condition of equilibrium. It is valid for translational (linear) motion

On the other hand, if the system tends to rotate around a fixed point, the equilibrium is settled only when the resultant moment of force is zero, i.e. when

MR = M1 + M2 + M3 + … = 0

Or

MR = F1 × ∆x1 + F2 × ∆x2 + F3 × ∆x3 + … = 0

The above equation is known as the second condition of equilibrium. It is valid for circular motion around a fixed point.

## Moment of Force. Conditions of Equilibrium Revision Questions

1) Two objects are placed on a massless bar in different distances from the turning point as shown in the figure below. What is the force F necessary to keep the system in equilibrium at the horizontal position?

1. 5.8 N
2. 3.4 N
3. 34 N
4. 58 N

2) A 80 N force acts on the left end of a 18 N heavy bar at 370 to the horizontal direction as shown in the figure. What is the magnitude of the extra weight placed on the right end of the bar necessary to set the equilibrium? Take cos 370 = 0.8 and sin 370 = 0.6.

1. 29 N
2. 14 N
3. 39.7 N
4. 34.8 N

3) What is m1 + m2 + m3 for the system shown in the figure if it is in equilibrium? The bars are massless. 1. 18 kg
2. 16 kg
3. 14 kg
4. 12 kg