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 = Fy / Fx
Fx = F cos θ
Fy = F sin θ
θ = tan-1 = Fy / Fx
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 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 = Fy / Fx
θ = tan-1 = 6 / 6
θ = tan-1 (1)
θ = 45⁰ 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 = 30⁰
Fx = ? and Fy =?
Fx = F cos θ
Fx = 50 × cos 30
Fx = 50 N × 0.866 (˙.˙ cos 30⁰ = 0.866)
Fx = 43.3 N
and
Fy = F sin θ
Fy = 50 × 0.5 (˙.˙ sin 30⁰ = 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 = Fy / Fx
θ = tan-1 = 12 / 5
θ = tan-1 (2.4)
θ = 22.6⁰ 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 = 30⁰ (with x-axis)
Horizontal component of force Fx = 20 N
Force F =?
Fx = F cos θ
20 N = F cos 30⁰
20 N = F × 0.866 (˙.˙ cos 30⁰ = 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 = ?
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
