Table of Contents
Numerical Problems
4.1 Find the resultant of the following forces:

10 N along xaxis

6 N along yaxis and

4 N along negative xaxis. (8.5 N making 45^{⁰ }with xaxis)
Solution: F_{x} = Net force along xaxis = 10.4 = 6 N
F_{y} = Force along yaxis = 5 N
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 }= F_{y }/ F_{x}
_{ }θ = tan^{1 }= 6 / 6
θ = tan^{1 }(1)
θ = 45^{⁰} with xaxis
4.2 Find the perpendicular components of a force of 50 N making an angle of 30⁰ with xaxis. (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
= 50 N × 0.866 (˙.˙ cos 30^{⁰} = 0.866)
F_{x} = 43.3 N
Similarly, 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 xcomponent is 12 N and ycomponent is 5 N. (13 N making 22.6⁰ with xaxis)
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 xaxis
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
= 100 N × 0.1 m
= 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 xaxis)
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 = 23.09
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
= 50 N × (2 × 0.16)
= 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 = 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 = 0.133 × 100 = 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 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 = 200 / 5 = 40 N
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