2-2 The equilibrium of a system of coplanar forces: practice questions
Applied Mathematics — First: Statics · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 69 questions on this lesson, each with an explanation.
A uniform ladder rests with its lower end on rough horizontal ground and its upper end against a smooth vertical wall, in a vertical plane perpendicular to the wall. When the ladder is about to slip, the friction force on its lower end acts:
Ahorizontally towards the wall
Bhorizontally away from the wall
Cvertically upwards
Dalong the ladder towards its upper end
Show answer and explanation
Correct answer: (A) horizontally towards the wall
Explanation: About to slip, the lower end tends to move away from the wall, so friction (μR₁) acts horizontally towards the wall.
It balances the reaction of the smooth wall, which is horizontal and perpendicular to the wall: μR₁ = R₂.
Question 2Easy
A light beam of length 10 m rests on supports at its two ends. A single load of 200 N hangs at the midpoint. Find the reaction at each support.
A20 N and 20 N
B50 N and 150 N
C100 N and 100 N
D200 N and 0
Show answer and explanation
Correct answer: (C) 100 N and 100 N
Explanation: Step 1: Single load at the midpoint, so the reactions are equal.
Step 2: RA + RB = 200 N, each = 200 ÷ 2 = 100 N.
∴ Reaction at each support = 100 N.
Question 3Easy
A weight W hangs in equilibrium from two vertical strings whose tensions are 25 N and 35 N. Find W.
A30 N
B875 N
C60 N
D10 N
Show answer and explanation
Correct answer: (C) 60 N
Explanation: Step 1: The weight is held by the two vertical strings (ΣFy = 0).
Step 2: W = 25 + 35 = 60 N.
∴ The weight = 60 N.
Question 4Medium
Two people carry a light pole AB of length 3 m, holding it at the ends A and B. A load of 90 N hangs 1 m from A. Find the force borne by each person.
A60 N at A, 30 N at B
B90 N at A, 0 at B
C30 N at A, 60 N at B
D45 N at A, 45 N at B
Show answer and explanation
Correct answer: (A) 60 N at A, 30 N at B
Explanation: Step 1: Take moments about A: FB × 3 = 90 × 1 = 90.
Step 2: FB = 90 ÷ 3 = 30 N.
Step 3: ΣFy = 0: FA = 90 − 30 = 60 N.
∴ Force at A = 60 N, force at B = 30 N.
Question 5Medium
The distance between the wheels of a motorcycle is 1.5 m. Its weight, 200 kg.wt, acts on the vertical line through the midpoint between the wheels, and the rider's weight, 80 kg.wt, acts on the vertical line 0.6 m in front of the rear wheel. Find the reaction of the ground on each wheel.
Afront 132 kg.wt, rear 148 kg.wt
Bfront 148 kg.wt, rear 132 kg.wt
Cfront 140 kg.wt, rear 140 kg.wt
Dfront 132 kg.wt, rear 68 kg.wt
Show answer and explanation
Correct answer: (A) front 132 kg.wt, rear 148 kg.wt
Explanation: Step 1: The forces are vertical: weights 200 and 80 kg.wt down, reactions R1 (front) and R2 (rear) up.
Step 2: Moments about the rear wheel: R1 × 1.5 = 200 × 0.75 + 80 × 0.6 = 150 + 48 = 198, so R1 = 132 kg.wt.
Step 3: Sum of the vertical forces = 0: R1 + R2 = 280, so R2 = 148 kg.wt.
∴ Front wheel 132 kg.wt, rear wheel 148 kg.wt.
Question 6Medium
A light beam AB of length 4 m rests on supports at A and B. A load of 80 N hangs 3 m from A. Find RA and RB.
ARA = 60 N, RB = 20 N
BRA = 30 N, RB = 50 N
CRA = 40 N, RB = 40 N
DRA = 20 N, RB = 60 N
Show answer and explanation
Correct answer: (D) RA = 20 N, RB = 60 N
Explanation: Step 1: The load is 3 m from A (1 m from B). Take moments about B: RA × 4 = 80 × 1 = 80.
Step 2: RA = 80 ÷ 4 = 20 N.
Step 3: ΣFy = 0: RB = 80 − 20 = 60 N.
∴ RA = 20 N and RB = 60 N.
Question 7Medium
A light rod AB of length 4 m is hinged at A and carries a load of 60 N at B. It is kept horizontal by a vertical force F applied 1 m from A. Find F.
A240 N
B15 N
C180 N
D60 N
Show answer and explanation
Correct answer: (A) 240 N
Explanation: Step 1: Take moments about A: F × 1 = 60 × 4.
Step 2: F = 240 ÷ 1 = 240 N.
∴ F = 240 N.
Question 8Hard
A uniform rod AB of length 2 m and weight 60 N is hinged at A to a vertical wall and kept horizontal by a light string joining B to a point C on the wall 1.5 m vertically above A. Find the magnitude of the reaction of the hinge.
A60 N
B40 N
C50 N
D30 N
Show answer and explanation
Correct answer: (C) 50 N
Explanation: Step 1: BC = 2.5 m, so the string makes an angle θ with the rod where sinθ = 0.6, cosθ = 0.8. Moments about A: T(0.6)(2) = 60 × 1 ⇒ T = 50 N.
Step 2: Hinge components: horizontal = T cosθ = 40 N, vertical = 60 − T sinθ = 30 N.
∴ R = √(40² + 30²) = 50 N.
Question 9Hard
A light beam AB of length 10 m rests on supports at C and D, where AC = 2 m and DB = 2 m. A load of 100 N hangs at end A. What load hung at end B makes the reaction at C zero (the beam about to tip about D)?
A400 N
B100 N
C800 N
D25 N
Show answer and explanation
Correct answer: (A) 400 N
Explanation: Step 1: When RC = 0 the beam is on the point of tipping about D; take moments about D.
Step 2: A is 8 m from D and B is 2 m from D: 100 × 8 = W × 2.
Step 3: 800 = 2W, W = 400 N.
∴ W = 400 N.
Question 10Hard
A uniform rod AB of weight 60 N is freely hinged at A and rests in equilibrium inclined at 60° to the horizontal under a horizontal force F acting at B. Find F.
A10√3 N
B20√3 N
C10 N
D30 N
Show answer and explanation
Correct answer: (A) 10√3 N
Explanation: Step 1: Let AB = 2l. Moments about A: F × (2l sin60°) = 60 × (l cos60°).
Step 2: F × 2(√3/2) = 60 × 1/2 ⇒ √3 F = 30.
∴ F = 30/√3 = 10√3 N.