Multiple Choice Questions
4.1 A particle is simultaneously acted upon by two forces of 4 and 3 newtons. The net force on the particle is:
(a) 1 N
(b) Between 1 N and 7 N
(c) 5 N
(d) 7 N
Correct Option: (c) 5 N
Reason: The resultant force is calculated using the Pythagoras theorem because the forces are perpendicular:
\[F_{\text{net}} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}.\]
4.2 A force \( F \) is making an angle of \( 60^\circ \) with the x-axis. Its y-component is equal to:
(a) \( F \)
(b) \( F \sin 60^\circ \)
(c) \( F \cos 60^\circ \)
(d) \( F \tan 60^\circ \)
Correct Option: (b) \( F \sin 60^\circ \)
Reason: The y-component of a force is given by \( F_y = F \sin \theta \), where \( \theta = 60^\circ \).
4.3 Moment of force is called:
(a) moment arm
(b) couple
(c) couple arm
(d) torque
Correct Option: (d) torque
Reason: The moment of force is referred to as torque, which is the product of force and the perpendicular distance from the axis of rotation.
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) \( \Sigma F = 0 \) and \( \Sigma \tau = 0 \)
(b) \( \Sigma F = 0 \) and \( \Sigma \tau \neq 0 \)
(c) \( \Sigma F \neq 0 \) and \( \Sigma \tau = 0 \)
(d) \( \Sigma F \neq 0 \) and \( \Sigma \tau \neq 0 \)
Correct Option: (a) \( \Sigma F = 0 \) and \( \Sigma \tau = 0 \)
Reason: For a body to be in equilibrium, the net force and the net torque acting on it must both be zero.
4.5 A shopkeeper sells his articles by a balance having unequal arms of the pans. 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) not certain
Correct Option: (a) loses
Reason: Placing the weights on the shorter arm requires a greater force to balance the torque, leading to a loss for the customer.
4.6 A man walks on a tight rope. He balances himself by holding a bamboo stick horizontally. It 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 Option: (c) Principle of moments
Reason: The man balances himself by distributing his weight and using the stick to adjust the torque, based on the principle of moments.
4.7 In stable equilibrium the centre of gravity of the body lies:
(a) At the highest position
(b) At the lowest position
(c) At any position
(d) Outside the body
Correct Option: (b) At the lowest position
Reason: In stable equilibrium, the body always tends to return to its original position, which is possible only if the center of gravity is at the lowest position.
4.8 The centre of mass of a body:
(a) Lies always inside the body
(b) Lies always outside the body
(c) Lies always on the surface of the body
(d) May lie within, outside, or on the surface
Correct Option: (d) May lie within, outside, or on the surface
Reason: The center of mass is the point at which the mass of a body is considered to be concentrated. It depends on the distribution of mass and may lie inside, outside, or on the surface of the object (e.g., a hollow ring’s center of mass lies outside the material).
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 Option: (a) Stable equilibrium
Reason: A cylinder resting on its circular base is in stable equilibrium because if it is tilted slightly, it will return to its original position due to its low center of gravity.
4.10 Centripetal force is given by:
(a) \( rF \)
(b) \( rF \cos \theta \)
(c) \( \frac{mv^2}{r} \)
(d) \( \frac{mv}{r} \)
Correct Option: (c) \( \frac{mv^2}{r} \)
Reason: Centripetal force is the force required to keep an object moving in a circular path and is given by the formula:
\[F = \frac{mv^2}{r}\]
where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the circular path.
Short Questions
4.1 Define like and unlike parallel forces.
The define of the like and unlike parallel forces are
Like parallel forces
Forces that act in the same direction along parallel lines.
Unlike parallel forces
Forces that act in opposite directions along parallel lines.
4.2 What are rectangular components of a vector and their values?
Rectangular components of a vector are the two perpendicular components of a vector along the x-axis and y-axis.
\( V_x = V \cos \theta \): The component along the x-axis.
\( V_y = V \sin \theta \): The component along the y-axis.
4.3 What is the line of action of a force?
The line of action of a force is an imaginary straight line that extends along the direction of the force and passes through the point where the force is applied.
4.4 Define moment of a force. Prove that \( \tau = r F \sin \theta \), where \( \theta \) is the angle between \( r \) and \( F \).
The moment of a force (torque) is the turning effect of a force about a point or axis.
Formula
\[\tau = r \cdot F \cdot \sin \theta\]
Proof
The force \( F \) has a perpendicular component \( F_\perp = F \sin \theta \) at a distance \( r \).
Torque \( \tau \) is given by the product of force and perpendicular distance:
\[\tau = r \cdot F_\perp = r \cdot F \cdot \sin \theta\]
4.5 With the help of a diagram, show that the resultant force is zero but the resultant torque is not zero.
If two equal and opposite forces act on a rigid body at different points, the resultant force becomes zero because they cancel out.
However, the torques produced by the forces are in the same direction, so the resultant torque is not zero.
4.6 Identify the state of equilibrium in each case in the figure given below.
(a) Stable equilibrium:
The cone will return to its original position when tilted.
(b) Unstable equilibrium:
The ball will roll off when displaced slightly.
(c) Neutral equilibrium:
The cylinder will stay in the same position even if moved slightly.
4.7 Give an example of the body which is moving yet in equilibrium.
A car moving at constant speed on a straight road is in equilibrium because the net force and net torque acting on it are zero.
4.8 Define centre of mass and centre of gravity of a body.
Centre of mass:
The point where the entire mass of the body is considered to be concentrated for motion analysis.
Centre of gravity:
The point where the entire weight of the body acts.
4.9 What are two basic principles of stability physics which are applied in designing balancing toys and racing cars?
Lowering the center of gravity:
Objects with a low center of gravity are more stable.
Wider base area:
A wide base provides more stability by increasing the area of support.
4.10 How can you prove that the centripetal force always acts perpendicular to velocity?
Centripetal force causes circular motion and is directed toward the center of the circle.
The velocity of the object is tangential to the circular path.
Since the radius (force direction) and tangent (velocity direction) are perpendicular, the centripetal force is always perpendicular to the velocity.
Constructed Response Questions
4.1 A car travels at the same speed around two curves with different radii. For which radius the car experiences more centripetal force? Prove your answer.
The centripetal force is given by:
\[F = \frac{mv^2}{r}\]
where
\( m \) is the mass of the car,
\( v \) is the speed of the car,
\( r \) is the radius of the curve.
For the same speed \( v \), the centripetal force is inversely proportional to the radius \( r \).
A smaller radius \( r \) results in a larger centripetal force.
Conclusion:
The car experiences more centripetal force when traveling around the curve with a smaller radius.
4.2 A ripe mango does not normally fall from the tree. But when the branch of the tree is shaken, the mango falls down easily. Can you tell the reason?
When the branch is shaken:
The mango’s inertia resists the motion caused by the shake.
The force due to shaking creates a separation between the mango and the branch.
Once the connection between the mango and the branch is broken, gravity pulls the mango down.
Conclusion:
The mango falls due to inertia and the breaking of the support caused by the shaking.
4.3 Discuss the concepts of stability and centre of gravity in relation to objects toppling over. Provide an example where an object’s centre of gravity affects its stability, and explain how altering its base of support can influence stability.
Stability
An object is stable when its center of gravity is low and lies within its base of support.
Centre of Gravity
The point where the weight of the object appears to act.
Toppling
An object topples when its center of gravity moves outside its base of support.
Example
A tall, narrow object (e.g., a ladder) is less stable because its center of gravity is high. A wider base (e.g., a person standing with legs apart) increases stability.
Conclusion
Lowering the center of gravity and increasing the base of support improves stability.
4.4 Why an accelerated body cannot be considered in equilibrium?
Equilibrium occurs when
The net force acting on a body is zero (\( \Sigma F = 0 \)).
The net torque acting on a body is zero (\( \Sigma \tau = 0 \)).
In an accelerated body
A net force is present, which causes the acceleration.
Since \( \Sigma F \neq 0 \), the body is not in equilibrium.
Conclusion
An accelerated body cannot be in equilibrium because the forces acting on it are unbalanced.
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?
The taller box is more likely to tumble over because:
It has a higher center of gravity.
When the truck stops suddenly, the inertia of the taller box causes a larger moment (torque) about its base.
This torque can topple the taller box more easily compared to the shorter box.
Conclusion
A lower center of gravity makes the shorter box more stable during sudden stops.
Comprehensive Questions
4.1 Explain the principle of moments with an example.
The principle of moments states:
For a body in rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments.
Example
A seesaw is balanced when the product of force and distance from the pivot is equal on both sides:
\[F_1 \cdot d_1 = F_2 \cdot d_2\]
If a heavier person sits closer to the pivot and a lighter person sits farther, they can balance the seesaw.
4.2 Describe how you could determine the centre of gravity of an irregular shaped lamina experimentally.
Suspend the lamina from one point and let it hang freely.
Draw a vertical line from the suspension point using a plumb line.
Suspend the lamina from another point and draw another vertical line.
The point where the two lines intersect is the center of gravity.
4.3 State and explain two conditions of equilibrium.
Translational Equilibrium:
The net force acting on the body is zero (\( \Sigma F = 0 \)).
This prevents linear acceleration.
Rotational Equilibrium:
The net torque acting on the body is zero (\( \Sigma \tau = 0 \)).
This prevents rotational acceleration.
Conclusion
Both conditions must be satisfied for a body to remain in complete equilibrium.
4.4 How the stability of an object can be improved? Give a few examples to support your answer.
The stability of an object can be improved by:
Lowering the center of gravity:
Example
Heavy vehicles like trucks have a low center of gravity for better stability.
Increasing the base of support:
Example
A gymnast stands with legs apart to improve balance.
Adding more weight at the base:
Example
Tall towers have heavy bases to prevent toppling.
Conclusion
Stability depends on lowering the center of gravity and increasing the base of support.
Numerical Problems
4.1 A force of 200 N is acting on a cart at an angle of 30° with the horizontal direction. Find the x and y-components of the force.
To Find:
The x-component (\( F_x \)) and y-component (\( F_y \)) of the force.
Given:
Force \( F = 200 \, \text{N} \)
Angle \( \theta = 30^\circ \)
Find the x-component:
\[F_x = F \cos \theta\]
Substitute values:
\[F_x = 200 \cos 30^\circ = 200 \cdot 0.866 = 173.2 \, \text{N}\]
Find the y-component:
\[F_y = F \sin \theta\]
Substitute values:
\[F_y = 200 \sin 30^\circ = 200 \cdot 0.5 = 100 \, \text{N}\]
Final Answer
\( F_x = 173.2 \, \text{N}, F_y = 100 \, \text{N} \)
4.2 A force of 300 N is applied perpendicularly at the knob of a door to open it as shown in the given figure. If the knob is 1.2 m away from the hinge, what is the torque applied? Is it positive or negative torque?
To Find:
The torque (\( \tau \)) and its direction (positive or negative).
Given:
Force \( F = 300 \, \text{N} \)
Distance from hinge \( r = 1.2 \, \text{m} \)
Force applied is perpendicular, so \( \sin \theta = 1 \).
Calculate torque:
\[\tau = r \cdot F \cdot \sin \theta\]
Substitute values:
\[\tau = 1.2 \cdot 300 \cdot 1 = 360 \, \text{N·m}\]
Direction
The torque is positive because it causes counterclockwise rotation.
Answer:
\( \tau = 360 \, \text{N·m}, \text{positive torque} \)
4.3 Two weights are hanging from a metre rule at the positions as shown in the given figure. If the rule is balanced at its centre of gravity (C.G.), find the unknown weight \( w \).
To Find:
The unknown weight \( w \).
Given:
Weight 1: \( w \, \text{N} \), positioned at \( 40 \, \text{cm} \) left of C.G.
Weight 2: \( 4 \, \text{N} \), positioned \( 30 \, \text{cm} \) right of C.G.
For rotational equilibrium, the clockwise moment equals the anticlockwise moment:
\[\text{Clockwise Moment} = \text{Anticlockwise Moment}\]
Write the equation for moments:
\[w \cdot 40 = 4 \cdot 30\]
Solve for \( w \)
\[w = \frac{4 \cdot 30}{40} = \frac{120}{40} = 3 \, \text{N}\]
Answer:
\( w = 3 \, \text{N} \)
4.4 A see-saw 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 see-saw is in equilibrium.
To Find:
The moments on each side and whether the see-saw is in equilibrium.
Given:
Child A
Weight \( W_A = 30 \cdot 9.8 = 294 \, \text{N} \), Distance \( d_A = 2 \, \text{m} \).
Child B
Weight \( W_B = 40 \cdot 9.8 = 392 \, \text{N} \), Distance \( d_B = 1.5 \, \text{m} \).
Calculate moment for Child A:
\[\text{Moment}_{A} = W_A \cdot d_A = 294 \cdot 2 = 588 \, \text{N·m}\]
Calculate moment for Child B
\[\text{Moment}_{B} = W_B \cdot d_B = 392 \cdot 1.5 = 588 \, \text{N·m}\]
Check equilibrium:
The moments are equal on both sides (\( 588 = 588 \)). Thus, the see-saw is in equilibrium.
Answer
The moments on each side are \( 588 \, \text{N·m} \), and the see-saw is in equilibrium.
4.5 A crowbar is used to lift a box as shown in the given figure. If the 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.
To Find:
The weight the other end bears.
Given:
Downward force \( F = 250 \, \text{N} \),
Distance to downward force \( d_1 = 30 \, \text{cm} = 0.3 \, \text{m} \),
Distance to upward force \( d_2 = 5 \, \text{cm} = 0.05 \, \text{m} \).
Torque equilibrium condition:
\[F \cdot d_1 = W \cdot d_2\]
Rearrange for \( W \):
\[W = \frac{F \cdot d_1}{d_2} = \frac{250 \cdot 0.3}{0.05} = 1500 \, \text{N}\]
Answer
The weight the other end bears is \( 1500 \, \text{N} \).
4.6 A 30 cm long spanner is used to open the nut of a car. If the torque required for it is 150 N·m, how much force \( F \) should be applied on the spanner?
To Find
The force \( F \) required.
Given:
Torque \( \tau = 150 \, \text{N·m} \),
Distance \( r = 30 \, \text{cm} = 0.3 \, \text{m} \).
Using the torque formula:
\[\tau = F \cdot r\]
Rearrange for \( F \):
\[F = \frac{\tau}{r} = \frac{150}{0.3} = 500 \, \text{N}\]
Answer
The force required is \( 500 \, \text{N} \).
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° with the ceiling, as shown in the figure. Determine the magnitude of force \( F \) and tension \( T \) in the string.
To Find:
The force \( F \) and tension \( T \).
Given:
Weight of the ball \( W = 5 \, \text{N} \),
Angle \( \theta = 60^\circ \).
Vertical force balance:
The vertical component of tension balances the weight:
\[T \cos \theta = W\]
Rearrange for \( T \):
\[T = \frac{W}{\cos \theta} = \frac{5}{\cos 60^\circ} = \frac{5}{0.5} = 10 \, \text{N}\]
Horizontal force balance:
The horizontal force \( F \) is balanced by the horizontal component of tension:
\[F = T \sin \theta\]
Substitute \( T = 10 \):
\[F = 10 \cdot \sin 60^\circ = 10 \cdot 0.866 = 8.66 \, \text{N}\]
Answer:
\( F = 8.66 \, \text{N}, T = 10 \, \text{N} \).
4.8 A signboard is suspended by means of two steel wires as shown in the figure. If the weight of the board is 200 N, what is the tension in the strings?
To Find:
The tension in each string.
Given:
Weight of the board \( W = 200 \, \text{N} \),
The board is symmetric, so the tension is evenly divided:
Vertical force balance:
The total tension balances the weight of the board:
\[T_1 + T_2 = W\]
Since the board is symmetric:
\[T_1 = T_2 = \frac{W}{2} = \frac{200}{2} = 100 \, \text{N}\]
Answer
The tension in each string is \( 100 \, \text{N} \).
4.9 One girl of 30 kg mass sits 1.6 m from the axis of a see-saw. Another girl of mass 40 kg wants to sit on the other side, so that the see-saw may remain in equilibrium. How far away from the axis should the second girl sit?
To Find:
The distance \( d_2 \) for the second girl.
Given:
Mass of first girl \( m_1 = 30 \, \text{kg} \), Distance \( d_1 = 1.6 \, \text{m} \),
Mass of second girl \( m_2 = 40 \, \text{kg} \).
For equilibrium:
\[m_1 \cdot d_1 = m_2 \cdot d_2\]
Rearrange for \( d_2 \):
\[d_2 = \frac{m_1 \cdot d_1}{m_2} = \frac{30 \cdot 1.6}{40} = 1.2 \, \text{m}\]
Answer
The second girl should sit \( 1.2 \, \text{m} \) away from the axis.
4.10 Find the tension in each string of the figure if the block weighs 150 N.
To Find:
The tensions \( T_1 \) and \( T_2 \).
Given:
Weight \( W = 150 \, \text{N} \),
Angles: \( \theta_1 = 60^\circ \), \( \theta_2 = 30^\circ \).
Vertical force balance:
The vertical components of the tensions balance the weight:
\[T_1 \sin \theta_1 + T_2 \sin \theta_2 = W\]
Substitute \( \sin 60^\circ = 0.866 \), \( \sin 30^\circ = 0.5 \):
\[T_1 \cdot 0.866 + T_2 \cdot 0.5 = 150 \tag{1}\]
Horizontal force balance:
The horizontal components of the tensions are equal:
\[T_1 \cos \theta_1 = T_2 \cos \theta_2\]
Substitute \( \cos 60^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \):
\[T_1 \cdot 0.5 = T_2 \cdot 0.866\]
\[T_1 = \frac{T_2 \cdot 0.866}{0.5} = 1.732 \cdot T_2 \tag{2}\]
Solve equations (1) and (2):
Substitute \( T_1 = 1.732 \cdot T_2 \) into equation (1):
\[1.732 \cdot T_2 \cdot 0.866 + T_2 \cdot 0.5 = 150\]
\[1.5 \cdot T_2 + 0.5 \cdot T_2 = 150\]
\[2 \cdot T_2 = 150 \]
\[T_2 = 75 \, \text{N}\]
Substitute \( T_2 = 75 \) into equation (2):
\[T_1 = 1.732 \cdot 75 = 129.9 \, \text{N} \, (\text{rounded to } 130 \, \text{N})\]
Answer:
\( T_1 = 130 \, \text{N}, T_2 = 75 \, \text{N} \).
