(a) Define the moment of a force. (1 mark)
____________________________
(b) A trapdoor has a hinge at end A, as shown in Fig. 1.1.
The trapdoor has length \(80\,\mathrm{cm}\) and weight \(75\,\mathrm{N}\). The mass of the trapdoor is uniformly distributed along its length.
A force \(F\) acts at right angles to the trapdoor at end B so that the trapdoor is held in equilibrium at an angle of \(42^\circ\) to the horizontal.
(i) State the principle of moments. (2 marks)
________________________________
(ii) Calculate the component of the weight that is perpendicular to the trapdoor. (1 mark)
component of weight = ______________________________ \(\mathrm{N}\)
(iii) Calculate the magnitude of the force \(F\). (2 marks)
\(F=\) ______________________________ \(\mathrm{N}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
(a) Definition of the moment of a force [1 mark]
The moment of a force about a point is the product of the force and the perpendicular distance from the line of action of the force to the point.
\(\mathrm{Moment}=F\times d_\perp\)
Answer: \( \boxed{\text{force}\times\text{perpendicular distance}} \)
(b)(i) Principle of moments [2 marks]
The trapdoor is in rotational equilibrium.
Therefore, the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point.
\(\sum \mathrm{clockwise\ moments}=\sum \mathrm{anticlockwise\ moments}\)
(b)(ii) Component of weight perpendicular to the trapdoor [1 mark]
The weight acts vertically downward. The trapdoor is at \(42^\circ\) to the horizontal, so the angle between the weight and the perpendicular direction to the trapdoor gives
\(\text{component of weight}=75\cos42^\circ\)
\(\text{component of weight}=55.7\,\mathrm{N}\)
Answer: \( \boxed{56\,\mathrm{N}} \)
(b)(iii) Magnitude of force \(F\) [2 marks]
Take moments about the hinge at A.
The force \(F\) acts at right angles to the trapdoor, so its perpendicular distance from A is \(80\,\mathrm{cm}\).
The trapdoor is uniform, so its weight acts at its centre, \(40\,\mathrm{cm}\) from A.
Using the component of weight perpendicular to the trapdoor:
\(F\times80=56\times40\)
\(F=\dfrac{56\times40}{80}\)
\(F=28\,\mathrm{N}\)
Answer: \( \boxed{28\,\mathrm{N}} \)
