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 coplanar couples act on a body: the first is formed by two forces of 8 N with an arm of 5 cm and acts anticlockwise, and the second by two forces of 6 N with an arm of 4 cm and acts clockwise. Find the resultant couple.
One couple is formed by forces of 10 N with an arm of 3 m acting anticlockwise; another by forces of 5 N with an arm of 2 m acting clockwise. Find the resultant couple.
A20 N·m clockwise
B50 N·m anticlockwise
C20 N·m anticlockwise
D40 N·m anticlockwise
Show answer and explanation
Correct answer: (C) 20 N·m anticlockwise
Explanation: Step 1: First couple: M = 10 × 3 = 30 N·m anticlockwise (+30).
Step 2: Second couple: M = 5 × 2 = 10 N·m clockwise (−10).
Step 3: Resultant = 30 − 10 = +20.
∴ The resultant couple = 20 N·m anticlockwise.
Question 3Medium
ABCD is a rectangle in which AB = 4 cm and BC = 6 cm. Forces of 10, 15, 10 and 15 newtons act along AB, BC, CD and DA respectively. The system is equivalent to
Aa couple of moment 120 N·cm
Ba couple of moment 60 N·cm
Ca single force of 50 N
Dzero (equilibrium)
Show answer and explanation
Correct answer: (A) a couple of moment 120 N·cm
Explanation: Step 1: The resultant vanishes: the 10 N forces along AB and CD are opposite, and so are the 15 N forces along BC and DA.
Step 2: 10 N along AB and CD: a couple of 10 × 6 = 60 N·cm.
Step 3: 15 N along BC and DA: a couple of 15 × 4 = 60 N·cm in the same sense (all forces are taken in one cyclic order).
Step 4: Total = 60 + 60 = 120 N·cm (also = twice the area 24 cm² × 2.5 N per cm).
∴ The system is equivalent to a couple of moment 120 N·cm.
Question 4Medium
A body is acted on by two couples: 20 N·m clockwise and 12 N·m anticlockwise. Find the moment and sense of the resultant couple.
A8 N·m anticlockwise
B32 N·m clockwise
C8 N·m clockwise
D240 N·m clockwise
Show answer and explanation
Correct answer: (C) 8 N·m clockwise
Explanation: Step 1: Take anticlockwise as positive: +12 and −20.
Step 2: Sum = 12 − 20 = −8.
Step 3: The result is negative, so the sense is clockwise.
∴ The resultant couple = 8 N·m clockwise.
Question 5Hard
A force F = 20ĵ N acts at the point A(3, 0) (metres), and a couple of 15 N·m clockwise also acts on the body. The force is replaced by an equal force at the origin O together with a couple. Find the total moment of the couple of the resulting system about O (anticlockwise positive).
A45 N·m clockwise
B45 N·m anticlockwise
C75 N·m anticlockwise
D60 N·m anticlockwise
Show answer and explanation
Correct answer: (B) 45 N·m anticlockwise
Explanation: Step 1: Moving F = 20ĵ from A(3, 0) to O adds a couple equal to the moment of F about O: M = x·Fy − y·Fₓ = 3×20 − 0 = +60 N·m.
Step 2: The existing couple is 15 N·m clockwise = −15 N·m.
Step 3: Total couple = +60 + (−15) = +45 N·m.
∴ The resulting couple about O = 45 N·m anticlockwise.
Question 6Hard
ABCD is a square of side 10 cm. Forces of 6 N act along AB and CD, and forces of 8 N act along BC and DA (each in the order of the letters). Two equal and opposite forces acting at A and C perpendicular to AC keep the square in equilibrium. Find the magnitude of each.
A7√2 N
B14 N
C10√2 N
D7 N
Show answer and explanation
Correct answer: (A) 7√2 N
Explanation: Step 1: The given forces form two couples in the same sense: 6 × 10 + 8 × 10 = 140 N·cm.
Step 2: The balancing couple has arm AC = 10√2 cm: F × 10√2 = 140.
∴ F = 14/√2 = 7√2 N.
Question 7Hard
ABCD is an isosceles trapezium in which AD ∥ BC, AD = 6 cm, BC = 22 cm and AB = DC = 10 cm. Forces represented completely by AB, BC, CD and DA act on it, and the force along BC is 55 N. Find the magnitude of the moment of the couple to which the system is equivalent.
A210 N·cm
B560 N·cm
C420 N·cm
D840 N·cm
Show answer and explanation
Correct answer: (C) 420 N·cm
Explanation: Step 1: The forces are represented completely by the sides of the closed figure taken in one cyclic order, so the system is equivalent to a couple of moment = twice the area × (force per unit length).
Step 2: Force per cm = 55 ÷ 22 = 2.5 N.
Step 3: Each end overhang = (22 − 6) ÷ 2 = 8 cm, so the height = √(10² − 8²) = 6 cm; area = ½(6 + 22) × 6 = 84 cm².
Step 4: Moment = 2 × 84 × 2.5.
∴ The moment = 420 N·cm.
Question 8Hard
Forces of 6 N, 8 N and 10 N act along the sides AB, BC and CA of a triangle ABC (in the order of the letters), where AB = 3 cm, BC = 4 cm and CA = 5 cm. The system is equivalent to
Aa couple of moment 24 N·cm
Ba couple of moment 48 N·cm
Ca couple of moment 12 N·cm
Da single force of 24 N
Show answer and explanation
Correct answer: (A) a couple of moment 24 N·cm
Explanation: Step 1: The forces are 2 × the side lengths in cyclic order, so their vector sum is zero: the system is a couple.
Step 2: Moment about A: only the force along BC has a moment, with arm AB = 3 cm (angle B = 90°): 8 × 3 = 24 N·cm.
∴ A couple of moment 24 N·cm (= 2 × area × 2).
Question 9Hard
Forces F₁ = 2î − 3ĵ, F₂ = −î + 4ĵ and F₃ act at the points (1, 1), (−2, 0) and (0, 3) respectively, and the system is equivalent to a couple. Find the moment of the couple (anticlockwise positive, units consistent).
A5 units clockwise
B10 units anticlockwise
C16 units clockwise
D10 units clockwise
Show answer and explanation
Correct answer: (D) 10 units clockwise
Explanation: Step 1: A couple needs zero vector sum, so F₃ = −F₁ − F₂ = −î − ĵ.
Step 2: Moments about O (x·Fy − y·Fx): F₁: 1(−3) − 1(2) = −5; F₂: (−2)(4) − 0 = −8; F₃: 0 − 3(−1) = 3.
∴ Total = −10, i.e. 10 units clockwise (the same about every point).
Question 10Hard
In rectangle ABCD, A(0, 0), B(8, 0), C(8, 5), D(0, 5). Forces of 3 N act along AB (A→B) and CD (C→D); forces of 10 N act along AD (A→D) and CB (C→B). Find the resultant couple.
A65 N·m anticlockwise
B74 N·m clockwise
C95 N·m clockwise
D65 N·m clockwise
Show answer and explanation
Correct answer: (D) 65 N·m clockwise
Explanation: Step 1: Forces 3 N along AB and CD: arm = BC = 5 m, M = 3 × 5 = 15 N·m; A→B (right, bottom) and C→D (left, top) give anticlockwise (+15).
Step 2: Forces 10 N along AD and CB: arm = AB = 8 m, M = 10 × 8 = 80 N·m; A→D (up) on the left and C→B (down) on the right give clockwise (−80).
Step 3: Resultant = +15 + (−80) = −65 N·m.
∴ The resultant couple = 65 N·m clockwise.