# Class 9 Physics Chapter 4 Numerical Problems 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 = √Fx2 + Fy2

θ = tan-1 = F/ Fx

Fx = F cos θ

Fy = F sin θ

θ = tan-1 = F/ Fx

Τ = F × L

T – w = 0

F1 × L=  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 45⁰ with x-axis)

Solution:

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

Fy = Force along y-axis = 5 N

The magnitude of the resultant force = F =?

F = √Fx2 + Fy2

F = √(6)2 + (6)2

F = √36 + 36

= √72 = 8.5 N

As,

θ = tan-1 = F/ Fx

θ = tan-1 = 6 / 6

θ = tan-1 (1)

θ = 45 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

Fx = ? and Fy =?

Fx = F cos θ

Fx = 50 × cos 30

Fx = 50 N × 0.866                 (˙.˙ cos 30 = 0.866)

Fx = 43.3 N

and

Fy = F sin θ

Fy = 50 × 0.5                          (˙.˙ sin 30 = 0.5)

Fy = 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:

Fx = 12 N

Fy = 5 N

Magnitude of the force = F =?

Direction of the force = θ =?

F = √Fx2 + Fy2

F = √(12)2 + (5)2

F = √144+25

F = 13 N

(ii)      θ = tan-1 = F/ Fx

θ = tan-1 = 12 / 5

θ = tan-1 (2.4)

θ = 22.6 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 (with x-axis)

Horizontal component of force Fx = 20 N

Force F =?

Fx = F cos θ

20 N = F cos 30

20 N = F × 0.866                  (˙.˙ cos 30 = 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

∑ Fx = 0          and               ∑ Fy = 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 = ?

T= w1 + w2

T= Mg + mg

T= (M + m)g

T= (3+5) × 10

= 8 × 10

= 80 N

Also,

T= mg

T= 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 = F= 200 N

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

Length of the spanner to tighten the same nut:

Force = F2 = 150 N

Length = L= ?

Since              Τ1 = Τ2

F1 × L=  F2 × L2

200 × 0.1 = 150 × L2

20 = 150 × L2

L= 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 = L= 20 cm = 0.2 m

Moment arm of w2 = L= 50 cm = 0.5 m

Force required to balance the bar F2 =?

By applying the principle of moments:

Clockwise moments = anticlockwise moments

F1 × L=  F2 × L2

mg × L=  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

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