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 37 questions on this lesson, each with an explanation.
Two equal, opposite, parallel forces each of magnitude 10 N form a couple whose sense is anticlockwise. The perpendicular distance between their lines of action is 0.5 m. Find the moment of the couple.
A2.5 N·m anticlockwise
B10 N·m anticlockwise
C5 N·m anticlockwise
D20 N·m anticlockwise
Show answer and explanation
Correct answer: (C) 5 N·m anticlockwise
Explanation: Step 1: Moment of a couple = (magnitude of one force) × (perpendicular distance between the two lines of action).
Step 2: M = F × d = 10 × 0.5.
Step 3: M = 5 N·m, and the given sense is anticlockwise.
∴ The moment of the couple = 5 N·m anticlockwise.
Question 2Easy
A couple is a system of two forces that are:
Aequal in magnitude and acting in the same direction
Bunequal in magnitude and perpendicular to each other
Cequal in magnitude, opposite in direction, parallel, with different lines of action
Dequal in magnitude and acting along the same straight line
Show answer and explanation
Correct answer: (C) equal in magnitude, opposite in direction, parallel, with different lines of action
Explanation: Step 1: A couple is two forces equal in magnitude and opposite in direction.
Step 2: Their lines of action are parallel but not the same, separated by a perpendicular distance.
Step 3: Their resultant force is zero, yet they produce a turning effect.
∴ Equal in magnitude, opposite in direction, parallel, with different lines of action.
Question 3Easy
Which statement about the moment of a couple is correct?
AIt is zero about points lying on one of the two forces
BIt depends on the point about which it is taken
CIt equals the sum of the magnitudes of the two forces
DIt is the same about every point in the plane
Show answer and explanation
Correct answer: (D) It is the same about every point in the plane
Explanation: Step 1: A couple has zero resultant force, so its turning effect does not depend on the chosen point.
Step 2: Taking moments about any point gives the same value F × d.
∴ The moment of a couple is the same about every point in the plane.
Question 4Medium
A couple is formed by a force of 5 N acting vertically upward along the line x = 1 and a force of 5 N acting vertically downward along the line x = 4 (distances in metres). Find the moment of the couple.
A15 N·m anticlockwise
B5 N·m clockwise
C15 N·m clockwise
D20 N·m clockwise
Show answer and explanation
Correct answer: (C) 15 N·m clockwise
Explanation: Step 1: The two forces are equal (5 N), opposite, and parallel along vertical lines, so they form a couple.
Step 2: Perpendicular distance between the lines = 4 − 1 = 3 m.
Step 3: M = F × d = 5 × 3 = 15 N·m.
Step 4: The upward force is on the left and the downward force on the right, giving a clockwise sense.
∴ The moment of the couple = 15 N·m clockwise.
Question 5Medium
In rectangle ABCD, AB = 5 m and BC = 3 m. A force of 8 N acts along AB in the direction A→B, and an equal force of 8 N acts along CD in the direction C→D. Find the moment of the couple they form.
A64 N·m anticlockwise
B40 N·m anticlockwise
C24 N·m clockwise
D24 N·m anticlockwise
Show answer and explanation
Correct answer: (D) 24 N·m anticlockwise
Explanation: Step 1: The forces along AB and CD are equal (8 N), opposite, and parallel, so they form a couple.
Step 2: The perpendicular distance between lines AB and CD equals BC = 3 m.
Step 3: M = 8 × 3 = 24 N·m.
Step 4: The bottom force acts to the right and the top force to the left, giving an anticlockwise sense.
∴ The moment of the couple = 24 N·m anticlockwise.
Question 6Medium
A horizontal force of 15 N acts in the positive x-direction along a line that is 4 m above a point O. To replace it by an equal force acting through O together with a couple, find the moment of that couple.
A19 N·m clockwise
B60 N·m anticlockwise
C60 N·m clockwise
D3.75 N·m clockwise
Show answer and explanation
Correct answer: (C) 60 N·m clockwise
Explanation: Step 1: Moving the force to a parallel line through O requires an added couple equal to the moment of the force about O.
Step 2: The perpendicular distance from O to the line of action = 4 m.
Step 3: Couple moment = 15 × 4 = 60 N·m.
Step 4: A force pointing in the positive x-direction acting above O turns clockwise about O.
∴ The required couple = 60 N·m clockwise.
Question 7Medium
A couple is formed by two forces each 12 N with lines of action 2 m apart. It is replaced by an equivalent couple whose forces are 4 m apart. Find the magnitude of each new force.
A6 N
B48 N
C3 N
D24 N
Show answer and explanation
Correct answer: (A) 6 N
Explanation: Step 1: Moment of the original couple = 12 × 2 = 24 N·m.
Step 2: Equivalent couples have equal moments, so 24 = F × 4.
Step 3: F = 24 / 4 = 6 N.
∴ Each new force = 6 N.
Question 8Hard
A force F = 3î + 4ĵ (newtons) acts at A(1, 2) and the force −F acts at B(5, −1) (coordinates in metres). Find the perpendicular distance between the two forces of this couple.
A4 m
B5 m
C25 m
D3 m
Show answer and explanation
Correct answer: (B) 5 m
Explanation: Step 1: M = BA × F with BA = (−4, 3): M = (−4)(4) − (3)(3) = −25, so |M| = 25 N·m.
Step 2: ‖F‖ = 5 N, so the arm = 25 ÷ 5.
∴ The arm = 5 m (here BA happens to be perpendicular to F, so it equals AB).
Question 9Hard
The two forces of a couple act at points A and B where AB = 4 m, and each force makes an angle of 30° with the line AB. If the moment of the couple is 60 N·m, find the magnitude of each force.
A10√3 N
B30 N
C60 N
D15 N
Show answer and explanation
Correct answer: (B) 30 N
Explanation: Step 1: The arm is the perpendicular distance between the lines of action: d = AB sin30° = 4 × 1/2 = 2 m.
Step 2: F × d = 60 ⇒ F = 60 ÷ 2.
∴ F = 30 N; using AB cos30° = 2√3 m as the arm gives 10√3 N, which is wrong.
Question 10Hard
A force F = 6î + 8ĵ (in newtons) acts at the point A(2, 1), and a force −6î − 8ĵ acts at B(2, 4) (coordinates in metres). These forces form a couple. Find the moment of the couple, taking anticlockwise as positive and using M = Σ(x·Fy − y·Fₓ).
A2 N·m anticlockwise
B18 N·m clockwise
C30 N·m anticlockwise
D18 N·m anticlockwise
Show answer and explanation
Correct answer: (D) 18 N·m anticlockwise
Explanation: Step 1: ‖F‖ = √(6² + 8²) = 10 N; the second force is its opposite, so together they form a couple.
Step 2: Moment of the first force about O = x·Fy − y·Fₓ = 2×8 − 1×6 = 10.
Step 3: Moment of the second force about O = 2×(−8) − 4×(−6) = −16 + 24 = 8.
Step 4: Sum = 10 + 8 = 18 (positive).
∴ The moment of the couple = 18 N·m anticlockwise.