Class 9 Physics Notes 2026 – Chapter 4: Turning Effects of Forces

Punjab Board

A. Multiple Choice Questions

4.1. A particle is simultaneously acted upon by two forces of 4 N and 3 N. The net force on the particle is:

(a) 1 N
(b) Between 1 N and 7 N
(c) 5 N
(d) 7 N

Correct Answer: (b) Between 1 N and 7 N

Explanation: The magnitude of the resultant of two forces ranges from the difference of their magnitudes, \(|4-3|=1\text{ N}\), to their sum, \(4+3=7\text{ N}\), depending on the angle between them.

4.2. A force \(F\) makes an angle of \(60^\circ\) with the x-axis. Its y-component is equal to:

(a) \(F\)
(b) \(F\sin60^\circ\)
(c) \(F\cos60^\circ\)
(d) \(F\tan60^\circ\)

Correct Answer: (b) \(F\sin60^\circ\)

Explanation: The y-component of a vector making an angle \(\theta\) with the x-axis is given by:

\[ F_y=F\sin\theta \]

4.3. Moment of force is called:

(a) Moment arm
(b) Couple
(c) Couple arm
(d) Torque

Correct Answer: (d) Torque

Explanation: The turning effect of a force is called torque or moment of force.

4.4. If \(F_1\) and \(F_2\) are the forces acting on a body and \(\tau\) is the torque produced in it, the body will be completely in equilibrium when:

(a) \(\sum F=0\) and \(\sum\tau=0\)
(b) \(\sum F=0\) and \(\sum\tau\ne0\)
(c) \(\sum F\ne0\) and \(\sum\tau=0\)
(d) \(\sum F\ne0\) and \(\sum\tau\ne0\)

Correct Answer: (a) \(\sum F=0\) and \(\sum\tau=0\)

Explanation: Complete equilibrium requires both the net force and the net torque to be zero. This ensures translational and rotational equilibrium.

4.5. A shopkeeper sells his articles using a balance with unequal arms. If he puts the weights in the pan having the shorter arm, then the customer:

(a) Loses
(b) Gains
(c) Neither loses nor gains
(d) It is not certain

Correct Answer: (a) Loses

Explanation: According to the principle of moments,

\[ W_1L_1=W_2L_2 \]

If the standard weights are placed on the shorter arm, a greater weight of goods is required to balance them. The customer receives less goods for the price paid and therefore loses.

4.6. A man walks on a tightrope. He balances himself by holding a bamboo stick horizontally. This is an application of:

(a) Law of conservation of momentum
(b) Newton’s second law of motion
(c) Principle of moments
(d) Newton’s third law of motion

Correct Answer: (c) Principle of moments

Explanation: Holding a long bamboo stick helps the tightrope walker adjust the turning effects of forces and maintain rotational balance. It also helps lower the overall centre of gravity.

4.7. In stable equilibrium, the centre of gravity of a body lies:

(a) At the highest position
(b) At the lowest position
(c) At any position
(d) Outside the body

Correct Answer: (b) At the lowest position

Explanation: When a body in stable equilibrium is displaced slightly, its centre of gravity rises. The body tends to return to its original position, where its centre of gravity is at the lowest possible position.

4.8. The centre of mass of a body:

(a) Always lies inside the body
(b) Always lies outside the body
(c) Always lies on the surface of the body
(d) May lie within, outside, or on the surface

Correct Answer: (d) May lie within, outside, or on the surface

Explanation: For regular solid objects, the centre of mass usually lies inside the body. For hollow or ring-shaped objects, it may lie outside the physical material, such as at the geometric centre of a ring.

4.9. A cylinder resting on its circular base is in:

(a) Stable equilibrium
(b) Unstable equilibrium
(c) Neutral equilibrium
(d) None of these

Correct Answer: (a) Stable equilibrium

Explanation: If the cylinder is tilted slightly, its centre of gravity rises, and it tends to return to its original position on its base.

4.10. Centripetal force is given by:

(a) \(rF\)
(b) \(rF\cos\theta\)
(c) \(\frac{mv^2}{r}\)
(d) \(\frac{mv}{r^2}\)

Correct Answer: (c) \(\frac{mv^2}{r}\)

Explanation: The formula for centripetal force in circular motion is:

\[ F_c=\frac{mv^2}{r} \]

B. Short Answer Questions

4.1. Define like and unlike parallel forces.

Answer:

Like parallel forces: Forces that act along parallel lines and in the same direction.
Unlike parallel forces: Forces that act along parallel lines but in opposite directions.

4.2. What are rectangular components of a vector and what are their values?

Answer: Rectangular components are two mutually perpendicular components of a vector along the x- and y-axes.

For a vector \(F\) making an angle \(\theta\) with the x-axis:

\[ F_x=F\cos\theta \]

\[ F_y=F\sin\theta \]

4.3. What is the line of action of a force?

Answer: The line of action of a force is the infinite straight line along which the force vector acts.

4.4. Define moment of a force. Prove that \(\tau=rF\sin\theta\), where \(\theta\) is the angle between \(r\) and \(F\).

Answer: The moment of a force, also called torque, is the turning effect of a force about a pivot point.

Proof:

Torque is equal to the force multiplied by the perpendicular distance from the pivot to the line of action of the force.

The perpendicular distance is:

\[ d=r\sin\theta \]

Therefore,

\[ \tau=F\times d \]

\[ \tau=F(r\sin\theta) \]

\[ \boxed{\tau=rF\sin\theta} \]

4.5. With the help of a diagram, show that the resultant force is zero but the resultant torque is not zero.

Answer: This situation represents a couple. Two equal and opposite parallel forces acting at different points produce zero net force but a non-zero net torque.

\[ \sum F=0 \]

\[ \sum\tau\ne0 \]

Examples include turning a water tap or a steering wheel.

Note: The diagram referred to in the question was not included in the supplied source.

4.6. Identify the state of equilibrium in each case in the given figure.

Answer:

(a) Cone on its base: Stable equilibrium.
(b) Sphere on a curved surface or bowl: Neutral or stable equilibrium, depending on its placement.
(c) Cylinder resting on its curved side: Neutral equilibrium.

Note: The original figure was not included in the supplied source.

4.7. Give an example of a body which is moving yet in equilibrium.

Answer: A parachutist falling vertically downward at constant terminal velocity is in equilibrium because the net force and acceleration are zero. Similarly, a car moving at constant speed along a straight highway can be in equilibrium if the forces acting on it balance.

4.8. Define centre of mass and centre of gravity of a body.

Answer:

Centre of mass: The point where the entire mass of a body appears to be concentrated for translational motion.
Centre of gravity: The point where the entire weight of the body appears to act.

4.9. What are two basic principles of stability in physics applied in designing balancing toys and racing cars?

Answer:

Lower the centre of gravity as much as possible.
Increase the base area of support.

4.10. How can you prove that the centripetal force always acts perpendicular to velocity?

Answer: In uniform circular motion, the velocity vector is tangential to the circular path at any given point, while the centripetal force points towards the centre of the circle. The radius is perpendicular to the tangent; therefore, centripetal force and velocity are perpendicular.

C. Constructed Response Questions

4.1. A car travels at the same speed around two curves with different radii. For which radius does the car experience more centripetal force? Prove your answer.

Answer: Centripetal force is given by:

\[ F_c=\frac{mv^2}{r} \]

Since the mass \(m\) and speed \(v\) are constant:

\[ F_c\propto\frac{1}{r} \]

Therefore, centripetal force is inversely proportional to the radius. The car experiences more centripetal force on the curve with the smaller radius.

4.2. A ripe mango does not normally fall from a tree. But when the branch of the tree is shaken, the mango falls down easily. Can you tell the reason?

Answer: When the branch is shaken, it moves suddenly, but the mango tends to remain at rest due to its inertia. This relative motion strains the stem, causing the ripe mango to detach and fall under gravity.

4.3. Discuss the concepts of stability and centre of gravity in relation to objects toppling over.

Answer: An object topples when the vertical line drawn downward from its centre of gravity falls outside its base of support.

For example, a racing car is built low and with a wide base to reduce the risk of toppling during sharp turns. Lowering the centre of gravity and widening the base of support increase stability.

4.4. Why cannot an accelerated body be considered in equilibrium?

Answer: In translational equilibrium, the net force and linear acceleration are zero. In rotational equilibrium, the net torque and angular acceleration are zero. If a body has non-zero linear acceleration, a net unbalanced force is acting on it. Therefore, it cannot be in translational equilibrium.

4.5. Two boxes of the same weight but different heights are lying on the floor of a truck. If the truck makes a sudden stop, which box is more likely to tumble over? Why?

Answer: The taller box is more likely to tumble over because its centre of gravity is higher. During sudden braking, the turning effect associated with the box’s inertia can more easily cause the line of action of its weight to fall outside its base compared with the shorter box.

D. Comprehensive Questions

4.1. Explain the principle of moments with an example.

Answer: The principle of moments states that a body is in rotational equilibrium when the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot.

\[ \sum\tau=0 \]

For a balanced seesaw:

\[ W_1L_1=W_2L_2 \]

Here, \(W_1\) and \(W_2\) are the weights, while \(L_1\) and \(L_2\) are their distances from the pivot.

4.2. Describe how you could determine the centre of gravity of an irregularly shaped lamina experimentally.

Answer: The centre of gravity can be determined using the plumb-line method.

Suspend the irregular lamina freely from a pinhole near one corner.
Hang a plumb line from the same point.
Draw a vertical line along the direction of the plumb string.
Suspend the lamina from another point and draw a second vertical line.
Repeat from a third point if required.

The intersection of the lines gives the centre of gravity of the lamina.

4.3. State and explain the two conditions of equilibrium.

Answer:

1. First condition of equilibrium: The vector sum of all forces acting on a body must be zero.

\[ \sum F=0 \]

This ensures translational equilibrium.

2. Second condition of equilibrium: The sum of all torques acting on a body must be zero.

\[ \sum\tau=0 \]

This ensures rotational equilibrium.

4.4. How can the stability of an object be improved? Give a few examples to support your answer.

Answer: Stability can be improved by:

Lowering the centre of gravity.
Increasing the base area of support.

Examples include tightrope walkers holding a long pole to help maintain balance and double-decker buses having heavy components, such as engines, positioned low to improve stability.

E. Numerical Problems

4.1. A force of 200 N is acting on a cart at an angle of \(30^\circ\) with the horizontal direction. Find the x- and y-components of the force.

Answer:

Given:

Force, \(F=200\text{ N}\)
Angle, \(\theta=30^\circ\)

X-component:

\[ F_x=F\cos30^\circ \]

\[ F_x=200\times0.866 \]

\[ \boxed{F_x=173.2\text{ N}} \]

Y-component:

\[ F_y=F\sin30^\circ \]

\[ F_y=200\times0.5 \]

\[ \boxed{F_y=100\text{ N}} \]

4.2. A force of 300 N is applied perpendicularly at the knob of a door to open it. If the knob is 1.2 m away from the hinge, what is the torque applied? Is it positive or negative torque?

Answer:

Given:

Force, \(F=300\text{ N}\)
Distance from hinge, \(r=1.2\text{ m}\)
Angle, \(\theta=90^\circ\)

Using:

\[ \tau=rF\sin\theta \]

\[ \tau=1.2\times300\times\sin90^\circ \]

\[ \boxed{\tau=360\text{ N}\cdot\text{m}} \]

According to the supplied solution, the torque is positive and anticlockwise.

Note: The direction of torque depends on the direction in which the force turns the door. The referenced figure was not included, so the stated sign is retained from the source.

4.3. Two weights are hanging from a metre rule at the positions shown in the given figure. If the metre rule is balanced at its centre of gravity, find the unknown weight \(w\).

Answer:

Given:

Left-side weight, \(w\), at a distance of 40 cm
Right-side weight, 4 N, at a distance of 30 cm

By the principle of moments:

\[ w\times40=4\times30 \]

\[ 40w=120 \]

\[ w=\frac{120}{40} \]

\[ \boxed{w=3\text{ N}} \]

4.4. A seesaw is balanced with two children sitting near either end. Child A weighs 30 kg and sits 2 metres away from the pivot, while child B weighs 40 kg and sits 1.5 metres from the pivot. Calculate the total moment on each side and determine if the seesaw is in equilibrium.

Answer:

Moment of child A:

\[ \tau_A=30\times9.8\times2 \]

\[ \boxed{\tau_A=588\text{ N}\cdot\text{m}} \]

Moment of child B:

\[ \tau_B=40\times9.8\times1.5 \]

\[ \boxed{\tau_B=588\text{ N}\cdot\text{m}} \]

Both sides produce equal and opposite moments. Therefore, the seesaw is in rotational equilibrium.

4.5. A crowbar is used to lift a box. If a downward force of 250 N is applied at the end of the bar, how much weight does the other end bear? The crowbar itself has negligible weight.

Answer:

Given:

Load distance = 5 cm
Effort arm = 30 cm + 5 cm = 35 cm
Applied effort, \(F=250\text{ N}\)

Using the distances specified in the supplied solution, the torque balance is:

\[ \text{Load}\times5=250\times30 \]

\[ 5\times\text{Load}=7500 \]

\[ \text{Load}=\frac{7500}{5} \]

\[ \boxed{\text{Load}=1500\text{ N}} \]

Note: The source gives the effort arm as 35 cm but uses 30 cm in its torque equation. The answer above retains the source’s equation and result.

4.6. A 30 cm long spanner is used to open the nut of a car. If the torque required is 150 N m, how much force \(F\) should be applied on the spanner?

Answer:

Given:

Torque, \(\tau=150\text{ N}\cdot\text{m}\)
Length of spanner, \(r=30\text{ cm}=0.3\text{ m}\)

Assuming the force acts perpendicularly:

\[ \tau=Fr \]

\[ 150=F\times0.3 \]

\[ F=\frac{150}{0.3} \]

\[ \boxed{F=500\text{ N}} \]

4.7. A 5 N ball hanging from a rope is pulled to the right by a horizontal force \(F\). The rope makes an angle of \(60^\circ\) with the ceiling. Determine the magnitude of force \(F\) and tension \(T\) in the string.

Answer:

Given:

Weight, \(W=5\text{ N}\)
Angle with the ceiling, \(60^\circ\)
Angle with the vertical, \(30^\circ\)

Calculate tension:

For vertical equilibrium:

\[ T\cos30^\circ=5 \]

\[ T\times0.866=5 \]

\[ T=\frac{5}{0.866} \]

\[ \boxed{T\approx5.8\text{ N}} \]

Calculate horizontal force:

For horizontal equilibrium:

\[ F=T\sin30^\circ \]

\[ F=5.8\times0.5 \]

\[ \boxed{F\approx2.9\text{ N}} \]

4.8. A signboard is suspended by two steel wires as shown in the given figure. If the weight of the board is 200 N, what is the tension in the strings?

Answer:

Given:

Weight of signboard, \(W=200\text{ N}\)
The board is symmetrically suspended.

By symmetry:

\[ T_1+T_2=200 \]

Since the tensions are equal:

\[ T_1=T_2=\frac{200}{2} \]

\[ \boxed{T_1=T_2=100\text{ N}} \]

Note: This result assumes the two wires share the load equally, as stated in the supplied solution. The figure and wire angles were not provided.

4.9. One girl of 30 kg mass sits 1.6 m from the axis of a seesaw. Another girl of 40 kg wants to sit on the other side so that the seesaw remains in equilibrium. How far away from the axis should the other girl sit?

Answer:

Given:

First girl’s mass, \(m_1=30\text{ kg}\)
First girl’s distance, \(r_1=1.6\text{ m}\)
Second girl’s mass, \(m_2=40\text{ kg}\)
Second girl’s distance, \(r_2=?\)

For equilibrium, the moments must be equal:

\[ m_1r_1=m_2r_2 \]

\[ 30\times1.6=40r_2 \]

\[ 48=40r_2 \]

\[ r_2=\frac{48}{40} \]

\[ \boxed{r_2=1.2\text{ m}} \]

4.10. Find the tension in each string as shown in the given figure if the block weighs 150 N.

Answer:

Given:

Weight of block, \(W=150\text{ N}\)
Angle used for the first string, \(30^\circ\)

Calculate \(T_1\):

Using the vertical component:

\[ T_1\cos30^\circ=150 \]

\[ T_1\times0.866=150 \]

\[ T_1=\frac{150}{0.866} \]

\[ \boxed{T_1\approx173.2\text{ N}} \]

Calculate \(T_2\):

Using the horizontal component:

\[ T_2=T_1\sin30^\circ \]

\[ T_2=173.2\times0.5 \]

\[ \boxed{T_2=86.6\text{ N}} \]

Note: The figure showing the string arrangement was not included in the supplied source. The calculations above follow the angles and component relationships given in the source.

Notes by Sir Dilawar

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