## Statics – Moment Of A Force

Let the handle of a tube-well with O as the fulcrum be AB. Generally, the portion OA is used as the handle, i.e., the handle can turn about O, in a vertical plane. Obviously, the axis of rotation is at right angles to the plane. When part OA is pushed down, OB moves up and water rises through a piston connected at point B.

- From experience, we know that, if a force F is applied at O, the handle does not turn. It is also hard to lift water by applying force F at a little distance from O. But by applying force F at the endpoint A, the handle can easily be turned. Hence, it is not true that a body will always rotate on the application of a force on it.
- Rotation is possible only when the line of action of the force passes through a point which is at some distance from the axis of rotation. The rotational tendency increases with the increase in the perpendicular distance of the line of action of the force from the centre of rotation O.

The perpendicular distance of the line of action of the force, from the axis of rotation is called the arm of that force. In the figure shown, the length of the arms of the force F are 0, OL and OM. Hence, the rotation of a body depends on two factors:

**Read and Learn More: Class 11 Physics Notes**

- Magnitude (F) and
- Arm (d) of the applied force.

They together generate a physical quantity known as the moment of a force.

**Moment Of A Force Definition:** The force applied on a body and the perpendicular distance of the line of action of the force from the axis of rotation together, set up a rotational tendency of the body and is called the moment of the force about that axis of rotation.

- The product of the magnitude of the applied force (F) and the perpendicular distance (d) of the line of action of the force from the axis of rotation gives the magnitude of the moment of force, G = Fd.
- As shown in the figure, for different lines of action of F, the generated moments are 0, Fx OL and Fx OM respectively. As OM > OL > 0, the generated moment is the largest, when the line of action of F passes through A.
- The handle can easily turn when F acts at A. But the generated moment is zero when F acts at O.
- It is important to note that, a moment of force always refers to an axis of rotation. As a special case, when a particle rotates in a plane perpendicular to the axis of rotation, the point of intersection of the axis with that plane is sometimes called the point of rotation.
- Then the term, ‘moment of a force about a point of rotation’, may be used.
- A hinged door or window opens easily when pushed at a point farthest from the hinges. A door cannot be opened easily when we push on the hinges.

**Unit And Dimension Of Moment:** As defined, moment of a force = magnitude of the force x perpendicular distance of the line of action of the force from the axis of rotation. Hence, unit of moment = unit of force x unit of length.

Dimension of moment = dimension of force x dimension of length = MLT^{-2} · L = ML^{2}T^{-2}

**Algebraic Notation Of Moment Of A Force:** A number of forces acting on the same plane are called coplanar forces. Rotation, set up by the coplanar forces, also remains confined in that plane only. Suppose a body AB is able to turn about point O on a fixed plane (on the plane of the paper.

- The moment of the force F
_{1}is zero. - For F
_{2}, the rotation of the body AB is anticlockwise. As per convention, for anticlockwise rotation, the moment of a force is taken as positive. - For F
_{3}, the rotation is clockwise. Here, the moment of the force F_{3}is taken as negative.

Hence, the moments of coplanar forces can be expressed as ordinary positive or negative algebraical quantities.

**Algebraic Sum Of The Moments Of Force:** Let F_{1}, F_{2} and F_{3} be coplanar forces that act simultaneously on the body AB. Algebraic sum of moments on the body G = 0 + F_{2} x OM + (-F_{3} x ON) = F_{2} X OM – F_{3} X ON

If the sum

- Is zero, the body will not rotate,
- Is positive, the body turns anticlockwise
- Is negative, it turns clockwise.

If the resultant of F_{1}, F_{2} and F_{3} is R and s is the arm of R, then, G = R x s.

**Geometrical Representation Of Moments:** Let AB represent the magnitude, direction and line of action of the force F.

Moment of the force F about O = magnitude of F x perpendicular distance of O from the line of action

= AB x OC = 2 x (1/2 AB X OC) = 2 X area of ΔOAB

Hence, a moment of a force about a point is twice the area of the triangle formed by the endpoints of the force vector and the point about which the moment is taken.

**Geometrical Representation Of Moments Vector Form:** The moment of a force (G) is a vector quantity. Moment is represented in vector form as \(\vec{G}=\vec{r} \times \vec{F}\)

- The vector \(\vec{G}\) is illustrated. O is the point of rotation, i.e., the point of intersection of the axis of rotation, with the plane of rotation (the plane of the paper).
- P is the point of application of the force \(\vec{F}, \vec{r}=\overrightarrow{O P}\) = position vector of the point of application of the force, with respect to the point of rotation.
- By the definition of a cross product, the vector \(\vec{G}=\vec{r} \times \vec{F},\), is directed upwards from the plane of the paper. Hence, the torque always acts along the direction of the axis of rotation.

**Condition For Not Toppling Of A Moving Train Or Car On A Horizontal Circular Track:** Suppose a train or a car takes a turn along a circular horizontal path about the point O. The effective centripetal force acting on the car is \(\frac{m v^2}{r}\), where m = mass of the car, v = speed of the car, r = OD = radius of the circular path.

Since the frictional forces of the wheels A and B with the road (f_{1} and f_{2} respectively) supply the necessary centripetal force, \(f_1+f_2=\frac{m v^2}{r}\)….(1)

Again the sum of the normal reactions on the two wheels by the ground is equal to the weight of the car, i.e., R_{1} + R_{2} = mg….(2)

Suppose the distance between the two wheels of the car = 2 d, i.e., AD = DB = d and the centre of mass (G) of the car is at a height h above the ground, i.e., GD = h.

So, the condition of not toppling of the car is attained if the algebraic sum of the moments of the forces about G is zero,

i.e., \(R_2 \times B D-R_1 \times A D+f_1 \times D G+f_2 \times D G=0\)

or, \(R_2 d-R_1 d+\left(f_1+f_2\right) h=0\)

or, R_{2}d-R_{1}d+(f_{1}+f_{2})h = 0 …(3)

From equations (1) and (3) we get, \(R_1-R_2=\frac{m v^2 h}{r d}\)…(4)

Solving equations (2) and (4) we get, \(R_1=\frac{m}{2}\left(g+\frac{v^2 h}{r d}\right)\)…(5)

and \(R_2=\frac{m}{2}\left(g-\frac{v^2 h}{r d}\right)\)…(6)

It is clear that the value of R_{1} can never be zero. But with the increase in speed of the car, the value of R_{2} gradually decreases. For a speed v_{0}, if the value of R_{2} is zero then,

g = \(\frac{v_0^2 h}{r d} \text { or, } v_0^2=\frac{d r g}{h} \text { or, } v_0=\sqrt{\frac{d r g}{h}}\)…(7)

In this situation, as reaction R_{2} is zero, wheel B remains floating above the path. If the value of the speed exceeds v_{0}, the car will be overturned outwards with wheel A as the centre, i.e., the car will be overturned in the opposite direction of the centre of rotation O.

**Special Note:** Circular Motion we observe that \(v_m=\sqrt{\mu r g}\)

Comparing this equation with equation (7) from above, it can be inferred that

if \(\mu>\frac{d}{h} \text {, then } v_m>v_0\)

i.e., the car will topple before skidding.

Again, \(\text { if } \mu<\frac{d}{h} \text {, then } v_m<v_0\)

i.e., the car will skid before it topples.

Now, normally the coefficient of friction between the wheels of a car and the road or the wheels of a train and the railway tracks is around 0.8. If the magnitude of \(\frac{d}{h}\) is less, then even vehicles moving with a small value of speed along a horizontal circular track would show a tendency to topple.

This is the reason why the ratio of d and h is kept high, while designing vehicles. It is to be noted that d is half the distance between the two wheels (either front or rear) and h is the height of the centre of mass from the surface on which the vehicle is moving.

As \(\frac{d}{h}\) is built into the design of a vehicle, it generally does not overturn. In reality, vehicles do not topple before skidding in most of the cases.

## Statics – Moment Of A Force Numerical Examples

**Example 1. Masses of 2 kg and 5 kg are suspended at the two ends of a light rod of length 1.4 m. At which point should the rod be supported at to keep the system horizontal?**

**Solution:**

The rod is kept horizontal, and so does not have any rotational motion. Suppose the rod is supported at O.

Taking moments about the point O, 2 x OA – 5 x OB = 0

Let OA = x

∴ OB = (1.4-x)

Then, 2 x = 5(1.4 -x) or, 7x = 7 or, x = 1 m.

Hence, the rod is supported 1 m away from the 2 kg mass.

**Example 2. One end of a rope of length L is tied to a vertical pole. A man is pulling the rope with a constant force at its other end. Which point on the pole should the rope be tied at to uproot the pole most easily?**

**Solution:**

Suppose the rope is tied at point B of the pole OA and is pulled by a force F from the other end C. In order to uproot the pole, a rotational motion has to be set up about the base O of the pole. The moment of the force F about O sets up the rotation.

The moment = F x OD

= F x OB sinθ

= F x BC cosθ sinθ

= \(\frac{1}{2}\) F- L sin2θ (BC = L)

The moment will be maximum when sin 2θ = 1 or, θ = 45°.

Hence, OB = BCcosθ = L cos45° = \(\frac{L}{\sqrt{2}}\)

Hence, the rope should be tied at a height of \(\frac{L}{\sqrt{2}}\) from the ground.

**Example 3. There is an obstruction of height h in front of a wheel of radius r weighing W. What is the minimum horizontal force that is to be applied at the centre O of the wheel to overcome the obstruction? Given, h< r.**

**Solution:**

The wheel touches the obstacle at point B. Suppose a force F attempts to turn the wheel about point B. The weight of wheel W, on the other hand, resists this rotation. Hence to overcome the obstacle, the moment of the force F about B must be greater than the moment of W about B.

For F to be the minimum, F x BD = W x BC

Here, BD = OC = OA – AC = r-h

BC = \(\sqrt{O B^2-O C^2}=\sqrt{r^2-(r-h)^2}=\sqrt{2 r h-h^2}\)

Hence, F(r-h) = \(W \sqrt{2 r h-h^2}\)

or, F = \(\frac{W \sqrt{2 r h-h^2}}{r-h}\).

**Example 4. A horizontal force F is applied on a uniform sphere of radius r and weight W, and as a result, the sphere slides over the horizontal table. If the coefficient of friction between the sphere and the table is p > then show that h = \(r\left(1-\frac{\mu W}{F}\right)\)**

**Solution:**

The point of contact between the table and the sphere is Q. Here, the force of friction f acts along the surface of the table through the point Q opposite to the force F. The weight of the sphere W acts vertically downwards through the centre of the sphere. Since the sphere slides over the horizontal table, it has no rotational motion.

So, with respect to the centre O, f x r = F(r- h) or, μWx r = F(r-h)

h = \(r\left(1-\frac{\mu W}{F}\right)\)

**Example 5. A uniform iron rod of length 50 cm is bent, at right angles at 30 cm from one end, giving it an L-shape. The L-shaped rod is suspended freely from its point of bending. Find the angle that the 30 cm arm makes with the vertical at equilibrium.**

**Solution:**

Suppose the rod is bent at point B, AB = 20 cm, BC = 30 cm, and B is the point of suspension. Suppose BC, in equilibrium, makes an angle θ with the vertical.

Let the mass per unit length of the rod be w. So the mass of AB is 20 w and the mass of BC is 30 w. At equilibrium, the moments about point B due to the two masses, should be equal.

Let x and y be the perpendicular distances of the centres of mass of the arms AB and BC respectively, from the vertical line through B. The centre of mass is at the midpoint of each arm.

∴ 20 wx = 30 wy

or, 20 x 10 cosθ = 30 x 15 sinθ or, tanθ = \(\frac{4}{9}\)

or, θ = 24°.