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CIE iGCSE Co-Ordinated Science P1.5.2 Turning effect of forces Exam Style Questions Paper 2

CIE iGCSE Co-Ordinated Science Paper 2 - All Chapters

Question

A uniform metre rule rests on a pivot at the 50 cm mark. A load L is placed at the 30 cm mark and a load of 6.0 N is placed at the 80 cm mark. The arrangement is balanced.

What is the weight of load L?

A. 6.0 N
B. 9.0 N
C. 16 N
D. 24 N

▶️ Answer/Explanation
The distance of L from the pivot is \(50 – 30 = 20\text{ cm}\), and the distance of the 6.0 N load is \(80 – 50 = 30\text{ cm}\).
At equilibrium, clockwise moment = anticlockwise moment: \(L \times 20 = 6.0 \times 30\).
Solving: \(L = \frac{180}{20} = 9.0\text{ N}\).
✅ Answer: (B)

Question

The diagram shows a crane supporting a load of 6000 N. The horizontal distance between the load and the pivot is x.

The load is balanced about the pivot by a concrete block of mass 10 000 kg. The horizontal distance of the concrete block from the pivot is 2.0 m.

Gravitational field strength g is 10 N/kg.

What is the distance of x?

A. 1.2 m
B. 3.3 m
C. 12 m
D. 33 m

▶️ Answer/Explanation
The weight of the concrete block is \(10\,000\text{ kg} \times 10\text{ N/kg} = 100\,000\) N.
At balance, the clockwise moment equals the anticlockwise moment: \(6000 \times x = 100\,000 \times 2.0\).
Solving gives \(x = \frac{200\,000}{6000} \approx 33\) m.
✅ Answer: (D)

Question

The diagram shows a triangular sheet of metal with sides of length \(50\text{ cm}\), \(40\text{ cm}\) and \(30\text{ cm}\). The sheet is free to move about a pivot at the top corner, as shown.

A cord is attached to the bottom left corner of the sheet and pulled with a horizontal force of \(5.0\text{ N}\) to the left.

What is the moment of the \(5.0\text{ N}\) force about the pivot?

A. \(150\text{ Ncm}\)  
B. \(200\text{ Ncm}\)  
C. \(250\text{ Ncm}\)  
D. \(600\text{ Ncm}\)

▶️ Answer/Explanation
The force acts horizontally, so the relevant perpendicular distance from the pivot is the vertical side of the right-angled triangle, which is \(40\text{ cm}\).
\(\text{Moment} = F \times d = 5.0 \times 40 = 200\text{ Ncm}\).
✅ Answer: (B)

Question

A uniform beam has a mass of 12 kg and a length of 4.0 m. The beam rests on horizontal ground. One end of the beam is now raised from the ground by a vertical force F. The other end of the beam remains in contact with the ground and acts as a pivot.

The gravitational field strength g is 10 N/kg.

What is the value of F?

A. 6.0 N  
B. 24 N  
C. 60 N  
D. 240 N

▶️ Answer/Explanation
The beam’s weight, \(W = mg = 12 \times 10 = 120\text{ N}\), acts at its centre, 2.0 m from the pivot.
Taking moments about the pivot: \(F \times 4.0 = 120 \times 2.0\), so \(F = \frac{240}{4.0} = 60\text{ N}\).
✅ Answer: (C)

Question

The diagrams show uniform metre rulers each pivoted at the 50 cm mark. Different weights are placed on the rulers at different distances from the 0 cm end.

Which ruler rotates in a clockwise direction?

▶️ Answer/Explanation
Clockwise rotation occurs when the clockwise moment exceeds the anticlockwise moment about the pivot at 50 cm. For ruler 1, the 4 N weight at 30 cm from pivot creates 4×30=120 Ncm anticlockwise moment, and 2 N at 40 cm creates 2×40=80 Ncm clockwise moment, so it rotates anticlockwise. Ruler 1 gives 2 N at 10 cm = 20 Ncm (anticlockwise) and 6 N at 30 cm = 180 Ncm (clockwise), so it rotates clockwise.
✅ Answer: (A)
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