9th Class Physics · Punjab Board · Chapter 4

Chapter 4: Turning Effect of ForcesNumericals Solved Step by Step

All 10 numericals of 9th Class Physics Chapter 4, Turning Effect of Forces, solved step by step: Given Data, To Find, Formula, Calculation and Result, the way you write them in the exam. Punjab Board new syllabus.

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Chapter 4· Turning Effect of Forces

Numericals Solved Step by Step

Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.

Numerical 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.

Given Data

F = 200 N
θ = 30°

To Find

x-component Fx
y-component Fy

Formulas

Fx = F cos θ

Fy = F sin θ

Calculation

Fx = F cos θ = 200 × cos 30° = 200 × 0.866 = 173.2 N.
Fy = F sin θ = 200 × sin 30° = 200 × 0.5 = 100 N.

Result

Fx = 173.2 N, Fy = 100 N

Numerical 4.2

A force of 300 N is applied perpendicularly at the knob of a door to open it. If the knob is 1.2 m away from the hinge, what is the torque applied? Is it positive or negative torque?

The knob is 1.2 m from the hinge line, the axis of rotation.

Given Data

F = 300 N (perpendicular to the door)
moment arm l = 1.2 m

To Find

Torque τ, and its sign

Formula

τ = F × l

Calculation

The force is perpendicular, so the moment arm is the full 1.2 m.
τ = F × l = 300 × 1.2 = 360 N m.
In the book's figure the door turns anticlockwise, and anticlockwise torque is taken as positive.

Result

τ = 360 N m, positive (anticlockwise)

Numerical 4.3

Two weights hang from a metre rule that is balanced at its centre of gravity (C.G). An unknown weight w hangs 40 cm to the left of the C.G, and a 4 N weight hangs 30 cm to the right. Find w.

The rule balances on the wedge at its C.G.

Given Data

4 N weight at 30 cm = 0.3 m from the C.G (clockwise)
w at 40 cm = 0.4 m from the C.G (anticlockwise)

To Find

Unknown weight w

Formula

clockwise moments = anticlockwise moments

Calculation

By the principle of moments: 4 × 0.3 = w × 0.4.
1.2 = 0.4 w.
w = 1.2 ÷ 0.4 = 3 N.

Result

w = 3 N

Numerical 4.4

A see-saw is balanced with two children sitting near either end. Child A has a mass of 30 kg and sits 2 m from the pivot; child B has a mass of 40 kg and sits 1.5 m from the pivot. Calculate the moment on each side and decide whether the see-saw is in equilibrium. (g = 10 m s⁻²)

Given Data

m₁ = 30 kg, l₁ = 2 m
m₂ = 40 kg, l₂ = 1.5 m
g = 10 m s⁻²

To Find

Moment of each child
Is the see-saw in equilibrium?

Formulas

W = mg

τ = W × l

Calculation

Weights: W₁ = 30 × 10 = 300 N and W₂ = 40 × 10 = 400 N.
Moment of A: τ₁ = 300 × 2 = 600 N m (one way).
Moment of B: τ₂ = 400 × 1.5 = 600 N m (the other way).
The two moments are equal and opposite, so Στ = 0.

Result

Each moment is 600 N m, so the see-saw is in equilibrium

Note: The question's "weighs 30 kg" means a mass of 30 kg; weight is a force in newtons.

Numerical 4.5

A crowbar is used to lift a box. A downward force of 250 N is applied at the end of the bar, 30 cm from the pivot. The box rests on the other end, 5 cm from the pivot. How much weight does that end bear? The crowbar itself has negligible weight.

Effort 30 cm from the pivot, load 5 cm from it.

Given Data

F = 250 N
l₁ = 30 cm = 0.3 m
l₂ = 5 cm = 0.05 m

To Find

Load (weight) W on the short end

Formula

F × l₁ = W × l₂

Calculation

By the principle of moments about the pivot: 250 × 0.3 = W × 0.05.
75 = 0.05 W.
W = 75 ÷ 0.05 = 1500 N.

Result

W = 1500 N

Note: A 250 N push lifts 1500 N: the long arm is six times the short arm.

Numerical 4.6

A 30 cm long spanner is used to open the nut of a car. If the torque required is 150 N m, how much force should be applied at the end of the spanner?

F acts at right angles at the end of the spanner, so the moment arm l is the full 30 cm.

Given Data

l = 30 cm = 0.3 m
τ = 150 N m

To Find

Force F

Formula

τ = F × l, so F = τ ÷ l

Calculation

F = τ ÷ l = 150 ÷ 0.3.
F = 500 N.

Result

F = 500 N (applied at right angles to the spanner)

Numerical 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. Determine the force F and the tension T in the rope.

Three forces on the ball: T along the rope, F to the right, weight 5 N down.

Given Data

W = 5 N
rope at θ = 60° to the ceiling (horizontal)

To Find

Horizontal force F
Tension T

Formulas

ΣFy = 0: T sin θ = W

ΣFx = 0: F = T cos θ

Calculation

Resolve T: horizontal part T cos 60° (to the left), vertical part T sin 60° (up).
Vertical balance: T sin 60° = 5, so T = 5 ÷ 0.866 = 5.77 N.
Horizontal balance: F = T cos 60° = 5.77 × 0.5 = 2.89 N.

Result

T ≈ 5.77 N, F ≈ 2.89 N

Note: Some solved copies round T to 5.8 N first and get F = 2.9 N; the exact value is F = 5 ÷ tan 60° = 2.89 N.

Numerical 4.8

A 200 N signboard hangs from two vertical steel wires fixed 2 m on either side of its centre of gravity. Find the tension in each wire.

The wires are 2 m on either side of the board's centre of gravity.

Given Data

W = 200 N
each wire 2 m from the centre of gravity
wires vertical

To Find

Tensions T₁ and T₂

Formulas

ΣF = 0: T₁ + T₂ = W

Στ = 0 about G: T₁ × 2 = T₂ × 2

Calculation

Moments about G: T₁ × 2 = T₂ × 2, so T₁ = T₂.
Forces: T₁ + T₂ − 200 = 0, so 2T₁ = 200.
T₁ = T₂ = 100 N.

Result

T₁ = T₂ = 100 N

Numerical 4.9

A girl of mass 30 kg 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 stays in equilibrium. How far from the axis should she sit? (g = 10 m s⁻²)

Given Data

m₁ = 30 kg, l₁ = 1.6 m
m₂ = 40 kg
g = 10 m s⁻²

To Find

Distance l₂ of the second girl

Formula

m₁g × l₁ = m₂g × l₂

Calculation

By the principle of moments: 30 × 10 × 1.6 = 40 × 10 × l₂.
480 = 400 l₂.
l₂ = 480 ÷ 400 = 1.2 m.

Result

l₂ = 1.2 m from the axis

Note: g cancels from both sides, so 30 × 1.6 = 40 × l₂ gives the same answer.

Numerical 4.10

A 150 N block hangs from a knot. One string from the knot goes to the ceiling at 60° to it; a second string goes horizontally from the knot to a wall. Find the tension in each string.