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_{x}^{2} + F_{y}^{2}

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

F_{1} × L_{1 }=_{ }F_{2} × L_{2}

### 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_{x}^{2} + F_{y}^{2}

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_{x}^{2} + F_{y}^{2}

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 T_{1} = 3.8 N

Tension T_{2} = 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

(T_{1} + T_{2}) – w = 0

T_{1} + T_{2} = 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 = T_{1} = ? and T_{2} = ?

T_{1 }= w_{1} + w_{2}

T_{1 }= Mg + mg

T_{1 }= (M + m)g

T_{1 }= (3+5) × 10

= 8 × 10

= 80 N

Also,

T_{2 }= mg

T_{2 }= 3 × 10

T_{2} = 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_{1 }= 200 N

Length = L_{1 }= 10 cm = 10 / 100 = 0.1 m

Length of the spanner to tighten the same nut:

Force = F_{2} = 150 N

Length = L_{2 }= ?

Since Τ_{1} = Τ_{2}

F_{1} × L_{1 }=_{ }F_{2} × L_{2}

200 × 0.1 = 150 × L_{2}

20 = 150 × L_{2}

L_{2 }= 20 / 150 = 0.133 m

L_{2} = 0.133 × 100

L_{2} = 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 w_{1} = L_{1 }= 20 cm = 0.2 m

Moment arm of w_{2} = L_{2 }= 50 cm = 0.5 m

Force required to balance the bar F_{2} =?

By applying the principle of moments:

Clockwise moments = anticlockwise moments

F_{1} × L_{1 }=_{ }F_{2} × L_{2}

mg × L_{1 }=_{ }F_{2} × L_{2}

(10 ×10 ) × 0.2 =_{ }F_{2} × 0.5

20 = F_{2} × 0.5

F_{2} = 20 / 0.5

F_{2} = 200 / 5

F_{2} = 40 N

**Class 9 Physics Full Book Numerical Problems with Solution**