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
Comment
Good
Comment veryy good
Nice
Exelent
helpful
Very helpfull
That’s great for students to get better grades in class 8th and 9th. If they know the name of the website then it will be easy for them during preparations of exams. Those who don’t know about the website cannot do anything but to get help from parents and elder sisters and brothers.😚👌❤️😍🤪
It is great for students to get better grades in class 8th and 9th. This is really helpful in getting better grades and position.
Thank u