### Table of Contents

**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

Difficulty: Medium

**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

Now, θ = $tan^{-1}$ =$\frac{F_{y}}{F_{x}}$

θ = $tan^{-1}$ =$\frac{6}{6}$

θ = $tan^{-1}$ (1)

θ = 45⁰ with x-axis

**Find the perpendicular components of a force of 50 N making an angle of 30⁰ with an x-axis. (43.3 N, 25 N) **

Difficulty: Easy

**Solution:**

Force F = 50 N

Angle θ = 30⁰

$F_{x}$ = ? and $F_{y}$ = ?

$F_{x}$ = F cos θ

$F_{x}$= $50 \timescos 30$

= $50 N\times 0.866$ (˙.˙ cos 30⁰ = 0.866)

$F_{x}$ = 43.3 N

Similarly, $F_{y}$ = F sin θ

$F_{y}$= $50 \times 0.5$(˙.˙ sin 30⁰ = 0.5)

$F_{y}$= 25 N

**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)**

Difficulty: Medium

**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

**θ = $tan^{-1} =\frac{F_{y}}{F_{x}}$**

θ = $tan^{-1} =\frac{12}{5}$

θ = $tan^{-1}$ (2.4)

θ = 22.6⁰ with x-axis

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**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)**

Difficulty: Easy

**Solution:**

Force = F = 100 N

Distance = L = 10 cm = 0.1 m

Torque Τ =?

Torque Τ = $F \times L$

= 100 $N \times 0.1 m$

= 10 Nm

**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)**

Difficulty: Easy

**Solution:**

Angle θ = 30⁰ (with x-axis)

Horizontal component of force $F_{x}$ = 20 N

Force F = ?

$F_{x}$ = F cos θ

20 N = F cos 30⁰

20 N = $F \times 0.866$ (˙.˙ cos 30⁰ = 0.866)

F =$\frac{ 20 N}{0.866}$= 23.09

F = 23.1 N

**The steering of a car has a radius of 16 cm. Find the torque produced by a couple of 50 N. (16 Nm)**

Difficulty: Easy

**Solution:**

Radius = r = L = 16 cm =$\frac{16}{100}$ m = 0.16 m

Couple arm = L = 16 cm =$\frac{16}{100 }$ m = 0.16 m

Force = F = 50 N

Torque Τ = ?

Torque Τ = $F \times L$

= $50 N\times(2 \times 0.16)$

= 16 Nm

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**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)**

Difficulty: Easy

**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

**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)**

Difficulty: Medium

**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

$T_{1}$ = $(3+5) \times 10$

= $8 \times 10$

= 80 N

Also, $T_{2}$ = mg

$T_{2} = 3 \times10$ = 30 N

**A nut has been tightened by a force of 200 N using a 10 cm long spanner. What length of a spanner is required to loosen the same nut with 150 N force? (13.3 cm)**

Difficulty: Medium

**Solution:**

Force = $F_{1}$ = 200 N

Length = $L_{1}$ = 10 cm = $\frac{10 }{100}$= 0.1 m

Length of the spanner to tighten the same nut:

Force = $F_{2}$ = 150 N

Length = $L_{2}$ = ?

Since $Τ_{1} = Τ_{2}$

$F_{1} \times L_{1}$ = $F_{2} \times L_{2}$

$200 \times 0.1$ = $150 \times L_{2}$

20 = $150 \times L_{2}$

$L_{2}$ = $\frac{20}{150}$ = 0.133 m = $0.133 \times 100$ = 13.3 cm

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**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)**

Difficulty: Medium

**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

Did force require to balance the bar $F_{2}$ =?

**By applying principle of moments:**

Clockwise moments = anticlockwise moments

$F_{1} \times L_{1}$ = $F_{2} \times L_{2}$

$mg \times L_{1}$ = $F_{2} \times L_{2}$

$(10 \times10 )\times 0.2$ = $F_{2}\times0.5$

20 = $F_{2}\times 0.5$

$F_{2} =\frac{20}{0.5}$ = $\frac{200}{5}$= 40 N

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