M8.2, M8.4, M8.5Resolving forces and inclined planes
Edexcel A level Maths (9MA0) · Mechanics › Forces and Newton's laws
Practise Resolving forces and inclined planes. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A particle \(P\) of mass 1.2 kg is held in equilibrium on a smooth plane inclined at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{3}{4}\). It is held by a force of magnitude \(X\) newtons acting up the plane, along a line of greatest slope, as shown in the diagram. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
[object Object]
(a) Find the value of \(X\).[2]
(b) Find the magnitude of the normal reaction between \(P\) and the plane.[2]
Show the answer and mark scheme
(a)Answer: \(X = 7.06\) (accept 7.1)
M1 for resolving parallel to the plane: \(X = 1.2g \times \frac{3}{5}\)
A1 for 7.06 or 7.1
Worked solution: Parallel to the plane: \(X = 1.2g \times \frac{3}{5} = 7.06\)
(b)Answer: 9.41 N (accept 9.4 N) N
M1 for resolving perpendicular to the plane: \(R = 1.2g \times \frac{4}{5}\)
A1 for 9.41 or 9.4 (N)
Worked solution: Perpendicular to the plane: \(R = 1.2g \times \frac{4}{5} = 9.41\) N
Question 2Medium7 marks
A sledge of mass 15 kg is held at rest on a slope of snow by a rope. The slope is inclined at \(20^\circ\) to the horizontal and the rope is parallel to a line of greatest slope. The slope is modelled as a smooth plane, the sledge as a particle and the rope as light and inextensible. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
[object Object]
(a) Find the tension in the rope.[2]
(b) Find the magnitude of the normal reaction between the sledge and the slope.[2]
(c) A student says that the normal reaction should be equal to the weight of the sledge. Explain why the student is wrong.[1]
(d) In reality the snow is not smooth. State, with a reason, how this would affect the tension needed to hold the sledge at rest.[2]
Show the answer and mark scheme
(a)Answer: 50.3 N (accept 50 N) N
M1 for \(T = 15g\sin 20^\circ\)
A1 for 50.3 or 50 (N)
Worked solution: \(T = 15 \times 9.8 \times \sin 20^\circ = 50.3\) N (3 s.f.)
(b)Answer: 138 N (accept 140 N) N
M1 for \(R = 15g\cos 20^\circ\)
A1 for 138 or 140 (N)
Worked solution: \(R = 15 \times 9.8 \times \cos 20^\circ = 138\) N (3 s.f.)
(c)Answer: The normal reaction is perpendicular to the slope, so it only balances the component of the weight perpendicular to the slope, \(15g\cos 20^\circ\), which is less than the weight.
B1 for the reaction balances only the component of the weight perpendicular to the plane (\(mg\cos 20^\circ \lt mg\))
Worked solution: The reaction acts at right angles to the slope; the rope and the reaction together balance the weight, and the reaction only has to balance \(15g\cos 20^\circ\).
(d)Answer: The tension needed would be less (or could even be zero), because friction would act up the slope and help to hold the sledge.
B1 for the tension would be smaller (or it could be zero)
B1 for friction would act up the slope, opposing the tendency to slide down
Worked solution: Friction acts up the slope to oppose the sledge's tendency to slide down, so a smaller tension would be enough.
Question 3Hard10 marks
A particle \(P\) of weight \(W\) newtons rests in equilibrium on a smooth plane inclined at an angle \(\alpha\) to the horizontal. It is held by a horizontal force of magnitude 21 N, acting in the vertical plane containing a line of greatest slope. The normal reaction between \(P\) and the plane has magnitude 35 N.
[object Object]
(a) By resolving horizontally and vertically, show that \(\sin\alpha = \frac{3}{5}\).[3]
(b) Find the value of \(W\).[2]
(c) The horizontal force is replaced by a force of magnitude \(F\) newtons acting up the plane along a line of greatest slope, so that \(P\) remains in equilibrium. Find the value of \(F\) and the new magnitude of the normal reaction.[3]
(d) Show that, of all the forces that could hold \(P\) in equilibrium on the plane, the force acting up the line of greatest slope has the least magnitude.[2]
Show the answer and mark scheme
(a)Answer: \(35\sin\alpha = 21\), so \(\sin\alpha = \frac{3}{5}\) (shown)
M1 for resolving horizontally: the horizontal component of the reaction balances the 21 N force
A1 for \(35\sin\alpha = 21\)
A1* for \(\sin\alpha = \frac{3}{5}\)
Worked solution: The reaction is perpendicular to the plane, so it makes an angle \(\alpha\) with the vertical. Horizontally: \(35\sin\alpha = 21\), so \(\sin\alpha = \frac{21}{35} = \frac{3}{5}\).
(c)Answer: \(F = 16.8\) N and the normal reaction is 22.4 N
M1 for resolving parallel and perpendicular to the plane
A1 for \(F = 28 \times \frac{3}{5} = 16.8\)
A1 for \(R = 28 \times \frac{4}{5} = 22.4\)
Worked solution: Parallel: \(F = W\sin\alpha = 28 \times \frac{3}{5} = 16.8\) N. Perpendicular: \(R = W\cos\alpha = 28 \times \frac{4}{5} = 22.4\) N.
(d)Answer: For a force at angle \(\theta\) to the plane, \(F\cos\theta = W\sin\alpha\), so \(F = \frac{W\sin\alpha}{\cos\theta} \geqslant W\sin\alpha\), with equality when \(\theta = 0\).
M1 for resolving along the plane for a force at a general angle \(\theta\) to the plane: \(F\cos\theta = W\sin\alpha\)
A1* for \(F \geqslant W\sin\alpha\) with equality only when \(\theta = 0\) (force up the plane)
Worked solution: Only the component along the plane balances \(W\sin\alpha\): \(F\cos\theta = W\sin\alpha\). Since \(\cos\theta \leqslant 1\), \(F \geqslant W\sin\alpha = 16.8\), which is least when \(\theta = 0\).