M8.1, M8.4, M8.5Forces, resultants and equilibrium of a particle
Edexcel A level Maths (9MA0) · Mechanics › Forces and Newton's laws
Practise Forces, resultants and equilibrium of a particle. 3 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 1Easy6 marks
A particle \(P\) of mass 0.5 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(A\). A horizontal force of magnitude \(X\) N is applied to \(P\). The particle is in equilibrium with the string making an angle \(\theta\) with the vertical, where \(\tan \theta = \frac{3}{4}\), as shown in the diagram. The string and the force lie in the same vertical plane. 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 string.[3]
(b) Find the value of \(X\).[2]
(c) State how you have used the modelling assumption that the string is light.[1]
Show the answer and mark scheme
(a)Answer: 6.13 N (accept 6.1 N) N
M1 for resolving vertically: \(T \times \frac{4}{5} = 0.5g\)
A1 for a correct equation
A1 for 6.13 or 6.1 (N)
Worked solution: Resolving vertically: \(T \times \frac{4}{5} = 0.5 \times 9.8\), so \(T = 6.13\) N (3 s.f.)
(b)Answer: \(X = 3.68\) (accept 3.7)
M1 for resolving horizontally: \(X = T \times \frac{3}{5}\) with their \(T\), or \(X = 0.5g\tan \theta\)
A1 for 3.68 or 3.7
Worked solution: Resolving horizontally: \(X = T \times \frac{3}{5} = 3.68\)
(c)Answer: The weight of the string is ignored, so the string is straight and the tension is the same throughout the string.
B1 for the tension is the same throughout the string / the string has no weight so it stays straight and pulls along its length
Worked solution: A light string has negligible mass, so it hangs straight and has the same tension along its whole length; the force it exerts on \(P\) is along the string.
Question 2Medium9 marks
A shop sign of mass 12 kg hangs in equilibrium below a horizontal beam. The sign is supported by two cables, \(AP\) and \(BP\), which are attached to the beam at \(A\) and \(B\) and to the sign at the point \(P\). The cable \(AP\) is inclined at \(30^\circ\) to the horizontal and the cable \(BP\) is inclined at \(60^\circ\) to the horizontal, as shown in the diagram. The sign is modelled as a particle at \(P\) and the cables are modelled as light inextensible strings. 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) Show that the tension in the cable \(BP\) is 102 N, correct to 3 significant figures.[5]
(b) Find the tension in the cable \(AP\).[2]
(c) State what is meant by modelling the cables as light.[1]
(d) Give one reason why the answers found using this model may not be accurate.[1]
Show the answer and mark scheme
(a)Answer: \(T_B = 6\sqrt{3}g = 101.8\ldots \approx 102\) N (shown)
M1 for resolving horizontally with the correct number of terms (condone sign errors and sin/cos confusion)
A1 for \(T_A\cos 30^\circ = T_B\cos 60^\circ\)
M1 for resolving vertically with the correct number of terms (condone sign errors and sin/cos confusion)
A1 for \(T_A\sin 30^\circ + T_B\sin 60^\circ = 12g\)
A1* for solving to obtain \(T_B = 101.8\ldots\) and hence 102 N (3 s.f.)
Worked solution: Horizontally: \(T_A\cos 30^\circ = T_B\cos 60^\circ\), so \(T_A = \frac{T_B}{\sqrt{3}}\). Vertically: \(T_A\sin 30^\circ + T_B\sin 60^\circ = 12g\), so \(\frac{T_B}{2\sqrt{3}} + \frac{\sqrt{3}T_B}{2} = \frac{2T_B}{\sqrt{3}} = 12g\). \(T_B = 6\sqrt{3}g = 101.8\ldots\), which is 102 N to 3 significant figures.
(b)Answer: 58.8 N (accept 59 N or \(6g\)) N
M1 for substituting their \(T_B\) into either equation
A1 for 58.8 or 59 (N)
Worked solution: \(T_A = \frac{T_B}{\sqrt{3}} = 6g = 58.8\) N
(c)Answer: The mass (weight) of each cable is negligible, so each cable is straight and has the same tension throughout.
B1 for the cables have negligible mass / weight (so the tension is the same throughout each cable)
Worked solution: The weight of the cables is ignored, so each cable is straight and the tension is the same all along it.
(d)Answer: For example: the cables have weight, so they sag and the tension varies along them; or the sign is not a particle, so the cables are attached at different points.
B1 for a valid reason, e.g. the cables are not light (they sag / have weight); the sign has size, so the forces do not all act at one point; the cables may stretch; wind may act on the sign
Worked solution: Real cables have weight, so they are not straight lines with constant tension; also the sign is an extended object, so the cables may be attached at different points on it.
Question 3Hard9 marks
Three forces act on a particle.
(a) A student claims that forces of magnitudes 4 N, 6 N and 11 N could keep the particle in equilibrium. Explain why the student is wrong.[2]
(b) The particle is in equilibrium under forces of magnitudes 4 N, 6 N and \(F\) N. Find the range of possible values of \(F\).[2]
(c) Given that \(F = 8\), find the angle between the directions of the 4 N force and the 6 N force, giving your answer in degrees to one decimal place.[3]
(d) Find the angle between the directions of the 6 N force and the 8 N force, giving your answer in degrees to one decimal place.[2]
Show the answer and mark scheme
(a)Answer: The resultant of the 4 N and 6 N forces has magnitude at most 10 N (when they act in the same direction), so it can never balance a force of 11 N.
M1 for considering the greatest possible resultant of two of the forces (or the triangle of forces)
A1 for the resultant of 4 N and 6 N is at most 10 N, which is less than 11 N, so the forces cannot be in equilibrium
Worked solution: For equilibrium the third force must be equal and opposite to the resultant of the other two. The resultant of 4 N and 6 N lies between \(6 - 4 = 2\) N and \(6 + 4 = 10\) N, so it cannot be 11 N.
(b)Answer: \(2 \leqslant F \leqslant 10\)
M1 for identifying the least and greatest possible resultants of the 4 N and 6 N forces
A1 for \(2 \leqslant F \leqslant 10\)
Worked solution: \(F\) must equal the magnitude of the resultant of the other two forces, which varies from 2 N (opposite directions) to 10 N (same direction). So \(2 \leqslant F \leqslant 10\).
(c)Answer: \(75.5^\circ\) °
M1 for resolving (or the cosine rule) to find the resultant of the 4 N and 6 N forces in terms of the angle \(\alpha\) between them
Worked solution: The resultant of the 4 N and 6 N forces must have magnitude 8 N: \(16 + 36 + 48\cos\alpha = 64\), so \(\cos\alpha = \tfrac{1}{4}\) and \(\alpha = 75.5^\circ\).
(d)Answer: \(151.0^\circ\) °
M1 for finding the angle between the resultant of the 4 N and 6 N forces and the 6 N force (e.g. \(\cos\phi = \frac{6 + 4 \times \frac{1}{4}}{8} = \frac{7}{8}\)), or the sine rule in the triangle of forces
A1 for \(151.0^\circ\) (allow 151)
Worked solution: Take the 6 N force along the \(x\)-axis. The resultant of the 4 N and 6 N forces has components \(6 + 4\cos\alpha = 7\) and \(4\sin\alpha = \sqrt{15}\), making an angle \(\phi\) with the 6 N force where \(\cos\phi = \tfrac{7}{8}\), so \(\phi = 29.0^\circ\). The 8 N force acts in the opposite direction to this resultant, so the angle between it and the 6 N force is \(180^\circ - 29.0^\circ = 151.0^\circ\).