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 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.
Solution:
Force (F) = 200 N
Angle (θ) = 30° with horizontal
Using trigonometric relationships:
x-component
Fx = F cos(θ)
Fx = 200 × cos(30°)
Fx = 200 × 0.866
Fx = 173.2 N
y-component
Fy = F sin(θ)
Fy = 200 × sin(30°)
Fy = 200 × 0.5
Fy = 100 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?
Solution:
Force (F) = 300 N
Distance from hinge (r) = 1.2 m
Force is perpendicular to the door
Calculate torque:
Torque (τ) = F × r
τ = 300 N × 1.2 m
τ = 360 Nm
The torque is positive because the force creates a counterclockwise rotation when viewed from above.
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.
Solution:
Unknown weight (w) on left side at 40 cm from center
4 N weight on right side at 30 cm from center
Rule is balanced at center of gravity (C.G.)
Using the principle of moments:
Clockwise moment = Counterclockwise moment
w × 40 = 4 × 30
40w = 120
w = 3 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 sea-saw is in equilibrium.
Solution:
m1 = 30 kg,
h1 = 2 m from pivot
m2 = 40 kg,
h2 = 1.5 m from pivot
g = 9.8 m/s²
Calculate moments:
Moment A = 30 × 9.8 × 2 = 588 Nm
Moment B = 40 × 9.8 × 1.5 = 588 Nm
Total force = 30 × 9.8 = 294 N
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.
Solution:
Downward force = 250 N
Short arm = 5 cm
Long arm = 30 cm
Using the principle of moments:
Clockwise moment = Counterclockwise moment
250 × 30 = F × 5
F = (250 × 30)/5
F = 1500 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 as shown in the figure.
Solution:
Torque required τ = 150 Nm
Length r = 30 cm = 30/100 m = 0.3 m
Using formula,
τ = F × r
150 = F × 0.3
F = 150/0.3
F = 500 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 given figure. Determine the magnitude of force F and tension T in the string.
Solution:
Ball weight w = 5 N
Angle θ = 60°
Using force triangle:
F = wtanθ
F = 5 × tan(30°)
F = 2.9 N
T = w/cosθ
T = 5/cos(60°)
T = 5.8 N
4.8 A signboard is suspended by means of two steel 2 wires as shown in the given figure. If the weight of the board is 200 N, what is the tension in the strings?
Solution:
Weight = 200 N
Symmetric suspension with equal lengths
Due to symmetry and vertical equilibrium:
Each wire carries half the weight
T₁ = T₂ = 100 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, the other girl may sit?
Solution:
m1 = 30 kg
S1 = 1.6 m
m2 = 40 kg
distance S2 = ?
Using moment equilibrium:
F1S2 = F2S2
m1gS2 = m2gS2 (F = w = mg)
30 × 9.8 × 1.6 = 40 × 9.8 × S2
48 × 1.6 = 48S2
S2 = 1.2 m
4.10 Find the tension in each string of the as shown in the given figure, if the block weighs 150 N.
Solution:
Block weight (w) = 150 N
One string at angle (θ) = 60° from vertical
Second string is vertical
Step 1: Analyze forces
T₁ is the tension in the angled string (at 60°)
T₂ is the tension in the vertical string
Weight (w) = 150 N downward
Step 2: Set up equilibrium equations
For horizontal forces (x-direction):
T₁cos(30°) = 0 … (1)
For vertical forces (y-direction):
T₁cos(60°) + T₂ = 150 … (2)
Step 3: Solve for T₁
From equation (1):
T₁cos(30°) = T₁(0.866) = T₁sin(60°) = horizontal component This must equal horizontal component of T₂ to maintain equilibrium
T₁sin(60°) = T₁(0.866)
Step 4: Solve for T₁ using equation (2)
T₁(0.5) + T₂ = 150 … (3)
T₁(0.5) + 173.2 = 150
T₁(0.5) = -23.2
T₁ = 86.6 N
Step 5: Solve for T₂ From equation (3):
86.6(0.5) + T₂ = 150
43.3 + T₂ = 150
T₂ = 150 – 43.3
T₂ = 173.2 N
Therefore: T₁ = 86.6 N (tension in angled string) T₂ = 173.2 N (tension in vertical string)
Class 9 Physics Full Book Numerical Problems with Solution
