Q1: A pair of like parallel forces 15 N each are acting on a body. Find their resultant?
Given: Two like parallel forces F₁ = F₂ = 15 N
Formula: Resultant R = F₁ + F₂
Solution: R = 15 + 15 = 30 N
Answer: Resultant = 30 N
Formula: Resultant R = F₁ + F₂
Solution: R = 15 + 15 = 30 N
Answer: Resultant = 30 N
Q2: Two unlike parallel forces 10 N each acting along the same line. Find their resultant?
Given: Two unlike parallel forces F₁ = 10 N, F₂ = 10 N
Formula: Resultant R = |F₁ − F₂|
Solution: R = |10 − 10| = 0 N
Answer: Resultant = 0 N (they cancel each other)
Formula: Resultant R = |F₁ − F₂|
Solution: R = |10 − 10| = 0 N
Answer: Resultant = 0 N (they cancel each other)
Q3: Three forces 12 N along x-axis, 8 N making 45° with x-axis, and 8 N along y-axis
i) Find their resultant ii) Find the direction
i) Find their resultant ii) Find the direction
Components of 8 N at 45°:
Along x-axis: 8cos45° = 8 × 0.7071 ≈ 5.66 N
Along y-axis: 8sin45° = 8 × 0.7071 ≈ 5.66 N
Total x-component: 12 + 5.66 = 17.66 N
Total y-component: 8 + 5.66 = 13.66 N
i) Resultant R = √(x² + y²)
R = √(17.66² + 13.66²) = √(312.01 + 186.51) = √498.52 ≈ 22.33 N
ii) θ = tan⁻¹(y/x) = tan⁻¹(13.66 / 17.66) ≈ tan⁻¹(0.773) ≈ 37.8°
Along x-axis: 8cos45° = 8 × 0.7071 ≈ 5.66 N
Along y-axis: 8sin45° = 8 × 0.7071 ≈ 5.66 N
Total x-component: 12 + 5.66 = 17.66 N
Total y-component: 8 + 5.66 = 13.66 N
i) Resultant R = √(x² + y²)
R = √(17.66² + 13.66²) = √(312.01 + 186.51) = √498.52 ≈ 22.33 N
ii) θ = tan⁻¹(y/x) = tan⁻¹(13.66 / 17.66) ≈ tan⁻¹(0.773) ≈ 37.8°
Q4: A gardener is driving a lawnmower with a force of 80 N at 40° with ground
i) Find horizontal component ii) Find vertical component
i) Find horizontal component ii) Find vertical component
Given: F = 80 N, angle θ = 40°
i) Horizontal component: Fcosθ = 80 × cos(40°) ≈ 80 × 0.766 = 61.28 N
ii) Vertical component: Fsinθ = 80 × sin(40°) ≈ 80 × 0.643 = 51.44 N
i) Horizontal component: Fcosθ = 80 × cos(40°) ≈ 80 × 0.766 = 61.28 N
ii) Vertical component: Fsinθ = 80 × sin(40°) ≈ 80 × 0.643 = 51.44 N
Q5: Horizontal and vertical components of a force are 4 N and 3 N. Find
i) Resultant force ii) Direction of resultant
i) Resultant force ii) Direction of resultant
Given: Fx = 4 N, Fy = 3 N
i) Resultant R = √(Fx² + Fy²)
R = √(16 + 9) = √25 = 5 N
ii) Direction θ = tan⁻¹(Fy / Fx)
θ = tan⁻¹(3/4) ≈ 36.87°
i) Resultant R = √(Fx² + Fy²)
R = √(16 + 9) = √25 = 5 N
ii) Direction θ = tan⁻¹(Fy / Fx)
θ = tan⁻¹(3/4) ≈ 36.87°
Q6: A spanner of 0.3 m length produces torque of 300 Nm
i) Determine force applied
ii) What length for 500 Nm torque with same force?
i) Determine force applied
ii) What length for 500 Nm torque with same force?
Given: Torque τ = 300 Nm, r = 0.3 m
i) Formula: τ = r × F → F = τ / r = 300 / 0.3 = 1000 N
ii) New torque τ = 500 Nm, same F = 1000 N
r = τ / F = 500 / 1000 = 0.5 m
i) Formula: τ = r × F → F = τ / r = 300 / 0.3 = 1000 N
ii) New torque τ = 500 Nm, same F = 1000 N
r = τ / F = 500 / 1000 = 0.5 m
Q7: A uniform meter rule supported at center is balanced by 12 N and 20 N forces
i) 20 N is at 3 cm from pivot, find position of 12 N
ii) If 20 N is moved to 4 cm, find force to replace 12 N
i) 20 N is at 3 cm from pivot, find position of 12 N
ii) If 20 N is moved to 4 cm, find force to replace 12 N
i) Using clockwise torque = anticlockwise torque
20 × 3 = 12 × d → 60 = 12d → d = 60 / 12 = 5 cm
ii) New position of 20 N is 4 cm, let F be new force at 5 cm
F × 5 = 20 × 4 → F = 80 / 5 = 16 N
20 × 3 = 12 × d → 60 = 12d → d = 60 / 12 = 5 cm
ii) New position of 20 N is 4 cm, let F be new force at 5 cm
F × 5 = 20 × 4 → F = 80 / 5 = 16 N
Q8: a) Mechanic uses a double arm spanner to turn a nut. He applies a force of 15 N at each end of the spanner and produces a torque of 60 Nm. What is the length of the moment arm of the couple?
b) If he wants to produce a torque of 80 Nm with the same spanner then how much force should he apply?
b) If he wants to produce a torque of 80 Nm with the same spanner then how much force should he apply?
a) Formula for torque of a couple: τ = F × d
Given: Force at each end F = 15 N, τ = 60 Nm
d = τ / F = 60 / 15 = 4 m
b) Given: τ = 80 Nm, d = 4 m
F = τ / d = 80 / 4 = 20 N
Answer: a) 4 m b) 20 N
Given: Force at each end F = 15 N, τ = 60 Nm
d = τ / F = 60 / 15 = 4 m
b) Given: τ = 80 Nm, d = 4 m
F = τ / d = 80 / 4 = 20 N
Answer: a) 4 m b) 20 N
Q9: A uniform meter rule is balanced at the 30 cm mark when a load of 0.8 N is hung at the zero mark.
i) At what point on the rule is the center of gravity of the rule?
ii) Calculate the weight of the rule.
i) At what point on the rule is the center of gravity of the rule?
ii) Calculate the weight of the rule.
Let weight of the rule be W and CG at x cm from 0 mark
Clockwise moment: 0.8 × 30 = 24 Nm
Anticlockwise moment: W × (x − 30)
i) Rule is uniform → CG at center = 50 cm
W × (50 − 30) = 24 → W × 20 = 24 → W = 24 / 20 = 1.2 N
Answer: i) 50 cm ii) 1.2 N
Clockwise moment: 0.8 × 30 = 24 Nm
Anticlockwise moment: W × (x − 30)
i) Rule is uniform → CG at center = 50 cm
W × (50 − 30) = 24 → W × 20 = 24 → W = 24 / 20 = 1.2 N
Answer: i) 50 cm ii) 1.2 N
Q10: What will be moment of force when 500 N force is applied on a 40 cm long spanner to tighten a nut?
Given: Force = 500 N, Length = 40 cm = 0.4 m
Formula: Torque = F × r = 500 × 0.4 = 200 Nm
Answer: Torque = 200 Nm
Formula: Torque = F × r = 500 × 0.4 = 200 Nm
Answer: Torque = 200 Nm
Q11: If two forces 5 N each form a couple and the moment arm is 0.5 m, then what will be the torque of the couple?
Formula for torque of a couple: τ = F × d
Given: F = 5 N, d = 0.5 m
τ = 5 × 0.5 = 2.5 Nm
Answer: Torque = 2.5 Nm
Given: F = 5 N, d = 0.5 m
τ = 5 × 0.5 = 2.5 Nm
Answer: Torque = 2.5 Nm
Q12: A telephone pole of mass 300 kg is 30 m long. Its center of gravity is 10 m from the thick end.
What force must be applied at the thin end to keep it horizontal when supported at the midpoint?
What force must be applied at the thin end to keep it horizontal when supported at the midpoint?
Given: Mass = 300 kg → Weight = 300 × 9.8 = 2940 N
Pivot is at midpoint → 15 m from either end
CG is 10 m from thick end → 5 m from pivot (on thick side)
Let force F be applied at thin end (15 m from pivot)
Moment balance:
Clockwise: 2940 × 5 = 14700 Nm
Anticlockwise: F × 15
F × 15 = 14700 → F = 14700 / 15 = 980 N
Answer: Force = 980 N
Pivot is at midpoint → 15 m from either end
CG is 10 m from thick end → 5 m from pivot (on thick side)
Let force F be applied at thin end (15 m from pivot)
Moment balance:
Clockwise: 2940 × 5 = 14700 Nm
Anticlockwise: F × 15
F × 15 = 14700 → F = 14700 / 15 = 980 N
Answer: Force = 980 N
Q13: A uniform rod 10 meters long and weighing 30 N is supported in a horizontal position on a fulcrum with weights of 40 N and 50 N suspended from its ends as shown in the figure. Find the position of the fulcrum.
Let fulcrum be at x meters from the 40 N end.
Length of rod = 10 m
Weight of rod = 30 N acting at its center (5 m from either end)
Clockwise moments:
40 × x + 30 × (x − 5)
Anticlockwise moment:
50 × (10 − x)
Equating moments:
40x + 30x − 150 = 500 − 50x
120x = 650
x = 5.42 m
Answer: Fulcrum is 5.42 m from the 40 N end
Length of rod = 10 m
Weight of rod = 30 N acting at its center (5 m from either end)
Clockwise moments:
40 × x + 30 × (x − 5)
Anticlockwise moment:
50 × (10 − x)
Equating moments:
40x + 30x − 150 = 500 − 50x
120x = 650
x = 5.42 m
Answer: Fulcrum is 5.42 m from the 40 N end
Q14: A force of 25 N acts on a baby. If moment arm is 2 m, find the value of torque.
Formula: Torque = Force × Moment Arm
τ = 25 × 2 = 50 Nm
Answer: Torque = 50 Nm
τ = 25 × 2 = 50 Nm
Answer: Torque = 50 Nm
Q15: A force is applied perpendicularly on a door 4 meters wide which requires a torque of 120 Nm to open it. What will be the minimum force required?
Formula: Torque = Force × Distance → Force = Torque / Distance
F = 120 / 4 = 30 N
Answer: Minimum force required = 30 N
F = 120 / 4 = 30 N
Answer: Minimum force required = 30 N
Q16: What is the moment of the couple of 10 N acting at the extremities of a rod 5 m long? How can this couple be balanced?
Formula for couple torque: τ = Force × Distance between the forces
τ = 10 × 5 = 50 Nm
Balancing the couple: Apply an equal and opposite couple of 50 Nm in the opposite direction
Answer: Moment = 50 Nm. To balance, apply a couple of equal magnitude in the opposite direction.
τ = 10 × 5 = 50 Nm
Balancing the couple: Apply an equal and opposite couple of 50 Nm in the opposite direction
Answer: Moment = 50 Nm. To balance, apply a couple of equal magnitude in the opposite direction.
