Class 9 Physics Chapter 4 numerical problems are according to the new syllabus of the Punjab Board. Chapter 4 is related to the Turning Effect of Forces in which you will study force graphically and using different principles applied in daily life.

Class 9 Physics Chapter 4 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

F = √F_x² + F_y²

θ = tan^-1 = F_y / F_x

F_x = F cos θ

F_y = F sin θ

θ = tan^-1 = F_y / F_x

Τ = F × L

T – w = 0

F₁ × L₁ =  F₂ × L₂

Numerical Problems Chapter 4 – Turning Effect of Forces

4.1 Find the resultant of the following forces:

• 10 N along the x-axis
• 6 N along the y-axis and
• 4 N along the negative x-axis. (8.5 N making 45^⁰ with x-axis)

Solution:       

F_x = Net force along x-axis = 10.4 = 6 N

F_y = Force along y-axis = 5 N

The magnitude of the resultant force = F =?

F = √F_x² + F_y²

F = √(6)² + (6)²

F = √36 + 36

= √72 = 8.5 N

As,

θ = tan^-1 = F_y / F_x

θ = tan^-1 = 6 / 6

θ = tan^-1 (1)

θ = 45^⁰ with x-axis

4.2 Find the perpendicular components of a force of 50 N making an angle of 30⁰ with x-axis.               (43.3 N, 25 N)

Solution:       

Force = F = 50 N

Angle = θ = 30^⁰

F_x = ? and F_y =?

F_x = F cos θ

F_x = 50 × cos 30

F_x = 50 N × 0.866                 (˙.˙ cos 30^⁰ = 0.866)

F_x = 43.3 N

and    

F_y = F sin θ

F_y = 50 × 0.5                          (˙.˙ sin 30^⁰ = 0.5)

F_y = 25 N

4.3 Find the magnitude and direction of a force, if its x-component is 12 N and y-component is 5 N.                                                (13 N making 22.6⁰ with x-axis)

Solution:       

F_x = 12 N

F_y = 5 N

Magnitude of the force = F =?

Direction of the force = θ =?

F = √F_x² + F_y²

F = √(12)² + (5)²

F = √144+25

F = 13 N

(ii)      θ = tan^-1 = F_y / F_x

θ = tan^-1 = 12 / 5

θ = tan^-1 (2.4)

θ = 22.6^⁰ with x-axis

4.4 A force of 100 N is applied perpendicularly on a spanner at a distance of 10 cm from a nut. Find the torque produced by the force.                    (10 Nm)

Solution:        Force = F = 100 N

Distance = L = 10 cm = 0.1 m

Torque Τ = ?

Torque Τ = F × L

T = 100 N × 0.1 m

T = 10 Nm

4.5 A force is acting on a body making an angle of 30⁰ with the horizontal. The horizontal component of the force is 20 N. Find the force.            (23.1 N)

Solution:      

Angle θ = 30^⁰ (with x-axis)

Horizontal component of force F_x = 20 N

Force F =?

F_x = F cos θ

20 N = F cos 30^⁰

20 N = F × 0.866                  (˙.˙ cos 30^⁰ = 0.866)

F = 20 N / 0.866

F = 23.1 N

4.6 The steering of a car has a radius 16 cm. Find the torque produced by a couple of 50 N.                                                                             (16 Nm)

Solution:       

Radius = r = L = 16 cm = 16/100 m = 0.16 m

Couple arm = L = 16 cm = 16/100 m = 0.16 m

Force = F = 50 N

Torque = Τ =?

Torque = Τ = F × L

T= 50 N × (2 × 0.16)

T = 16 Nm

4.7 A picture frame is hanging by two vertical strings. The tensions in the strings are 3.8 N and 4.4 N. Find the weight of the picture frame.         (8.2 N)

Solution:       

Tension T₁ = 3.8 N

Tension T₂ = 4.4 N

Weight of the picture frame = w =?

When the picture is in equilibrium, then

∑ F_x = 0          and               ∑ F_y = 0

Therefore,

T – w = 0

Or

(T₁ + T₂) – w = 0

T₁ + T₂ = w

3.8 + 4.4 = w

w = 8.2 N

4.8 Two blocks of mass 5 kg and 3 kg are suspended by the two strings as shown. Find the tension in each string.                        (80 N, 30 N)

Solution:       

Mass of large block = M = 5 kg

Mass of large block = m = 3 kg

Tension produced in each string = T₁ = ? and T₂ = ?

T₁ = w₁ + w₂

T₁ = Mg + mg

T₁ = (M + m)g

T₁ = (3+5) × 10

= 8 × 10

= 80 N

Also,

T₂ = mg

T₂ = 3 × 10

T₂ = 30 N

4.9 A nut has been tightened by a force of 200 N using 10 cm long spanner. What length of a spanner is required to loosen the same nut with 150 N force?             (13.3 cm)

Solution:       

Force = F₁ = 200 N

Length = L₁ = 10 cm = 10 / 100 = 0.1 m

Length of the spanner to tighten the same nut:

Force = F₂ = 150 N

Length = L₂ = ?

Since              Τ₁ = Τ₂

F₁ × L₁ =  F₂ × L₂

200 × 0.1 = 150 × L₂

20 = 150 × L₂

L₂ = 20 / 150 = 0.133 m

L₂ = 0.133 × 100

L₂ = 13.3 cm

4.10 A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long. What force is required to balance it at its center of gravity by applying the force at the other end of the bar?                (40 N)

Solution:       

Mass of the block = m = 10 kg

Length of the bar = l = 1 m

Moment arm of w₁ = L₁ = 20 cm = 0.2 m

Moment arm of w₂ = L₂ = 50 cm = 0.5 m

Force required to balance the bar F₂ =?

By applying the principle of moments:

Clockwise moments = anticlockwise moments

F₁ × L₁ =  F₂ × L₂

mg × L₁ =  F₂ × L₂

(10 ×10 ) × 0.2 =  F₂ × 0.5

20 = F₂ × 0.5

F₂ = 20 / 0.5

F₂ = 200 / 5

F₂ = 40 N

Class 9 Physics Full Book Numerical Problems with Solution