Mechanics: Forces and Newton's Laws Depth
A-level Mechanics' forces and Newton's laws at depth combine resolving forces at angles, Newton's second law for single and connected particles, and friction, usually with an incline, a pulley or both in the same question. The marks reward a clear force diagram, resolving in sensible directions (parallel and perpendicular to an incline, rather than horizontal and vertical, whenever one is present), and correct treatment of a light inextensible string (tension equal throughout) and a smooth pulley (tension equal on both sides). Marks are commonly lost through sign errors in connected-particle systems, or by forgetting a force's component along or perpendicular to a slope.
Before you start
Make sure you're comfortable with these topics first:
Method
- Draw a force diagram for each object showing every force acting on it (weight, normal reaction, tension or thrust, friction, any applied force) - depth questions rarely state every force explicitly.
- Choose the resolving directions: use horizontal and vertical for objects on flat ground, but use directions parallel and perpendicular to the plane for an object on an incline, since this keeps the acceleration entirely within the parallel equation.
- Apply Newton's second law (F = ma) in the direction of acceleration, and equilibrium (resultant force = 0) in the perpendicular direction (or in every direction, if the object is not accelerating at all).
- For connected particles, write a separate F = ma equation for each particle, using the same letter for the acceleration's magnitude and the same letter for the tension in both equations (since the string is light and inextensible and any pulley is smooth), then solve the simultaneous equations.
- If friction is involved, decide whether the object is in limiting equilibrium (F = mu R exactly, on the point of sliding), moving (use F = mu R for kinetic friction, opposing the motion), or in general equilibrium (F <= mu R, so friction simply balances the other forces and cannot be found from mu R alone).
- Find the normal reaction R first, from the perpendicular (or vertical) equation, since R usually appears inside the friction term of the equation you actually need to solve.
- Keep the working algebraic for as long as possible, substituting numbers only at the end, so a 'show that' or 'find in terms of' part can still be answered from the same working.
Worked example
Three particles A, B and C, of masses 2 kg, 3 kg and 5 kg respectively, lie in that order on a rough horizontal floor and are joined in a line by two light inextensible strings, one connecting A to B and one connecting B to C. The coefficient of friction between each particle and the floor is 0.2. A horizontal force of 40 N is applied to C, pulling all three particles in the direction from A to C, starting from rest. Take g = 9.8 m/s^2. Find (a) the acceleration of the system, (b) the tension in the string joining A and B, (c) the tension in the string joining B and C.
- Treat the three particles as a single system to find the acceleration first: the driving force is 40 N, and friction acts on all three particles, totalling mu(m_A+m_B+m_C)g = 0.2(10)(9.8) = 19.6 N.
- Apply Newton's second law to the whole system: 40 - 19.6 = (2+3+5)a, so 20.4 = 10a, giving a = 2.04 m/s^2.
- Isolate particle A on its own to find the first tension: the only horizontal forces on A are the tension T1 pulling it forward and friction mu(m_A)g opposing it. So T1 - 0.2(2)(9.8) = 2a.
- Substitute a = 2.04: T1 - 3.92 = 2(2.04) = 4.08, so T1 = 8.00 N.
- Isolate particle B (or use C) to find the second tension: for B, the forces are T2 forward, T1 backward (Newton's third law reaction from the A-B string), and friction. So T2 - T1 - 0.2(3)(9.8) = 3a.
- Substitute T1 = 8 and a = 2.04: T2 - 8 - 5.88 = 3(2.04) = 6.12, so T2 = 20.0 N.
Practice questions
Try each question, then tap to reveal the answer.
Q1State Newton's second law of motion for a particle of constant mass.Show answer
Answer: The resultant force acting on a particle is equal to the particle's mass multiplied by its acceleration, F = ma, with F and a in the same direction (force in newtons, mass in kg, acceleration in m/s^2).
Q2A box of mass 2 kg rests in equilibrium on a rough horizontal floor. A horizontal force of 6 N is applied to the box, which remains in equilibrium. Find the frictional force acting on the box.Show answer
Answer: 6 N. Since the box is in equilibrium, friction exactly balances the applied force (there is no need to know mu, since the box is not on the point of sliding).
Q3A particle of mass 3 kg rests in equilibrium on a smooth inclined plane, held by a force of 12 N acting at 15 degrees above the line of greatest slope (i.e. not parallel to the plane). Take g = 9.8 m/s^2. Find the angle the plane makes with the horizontal, giving your answer to 3 significant figures.Show answer
Answer: 23.2 degrees (3 s.f.). Resolving along the plane: 12cos15 = 3g sin(theta), so sin(theta) = 12cos15/(3 x 9.8).
Q4A block of mass 5 kg rests on a rough horizontal floor. A force of 20 N is applied to the block at 25 degrees above the horizontal, pulling upward on the block. Take g = 9.8 m/s^2. Find the normal reaction between the block and the floor.Show answer
Answer: 40.5 N (3 s.f.), from R = 5g - 20sin25 (the upward component of the applied force reduces R).
Q5Two forces, (8i + 3j) N and (-2i + 5j) N, act simultaneously on a particle of mass 2 kg. Find the magnitude of the particle's acceleration, giving your answer as an exact value.Show answer
Answer: 5 m/s^2 (exact). The resultant force is (8-2)i + (3+5)j = 6i + 8j, with magnitude sqrt(36+64) = 10 N, so a = 10/2.
Q6Explain why, when two particles are connected by a string passing over a smooth pulley, the tension in the string is the same on both sides of the pulley.Show answer
Answer: The pulley is smooth (frictionless) and light (massless), so it exerts no resistive torque on the string passing over it and cannot change the magnitude of the tension; combined with the string itself being light, the tension is therefore the same throughout its length, on both sides of the pulley.
Q7A particle of mass 4 kg is on the point of sliding down a rough plane inclined at 15 degrees to the horizontal, under gravity alone. Find the coefficient of friction between the particle and the plane.Show answer
Answer: mu = tan15 = 0.268 (3 s.f.). At limiting equilibrium, friction acts up the plane and mg sin15 = mu(mg cos15), so mu = tan15 (the mass cancels).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A block of mass 8 kg rests on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A force of 50 N is applied to the block at an angle of 30 degrees above the horizontal, pulling the block along the ground. Take g = 9.8 m/s^2. (a) Find the normal reaction between the block and the ground. (3) (b) Show that the block accelerates, and find its acceleration. (4)
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Two particles, P of mass 5 kg and Q of mass 2 kg, are connected by a light inextensible string which passes over a smooth pulley fixed at the top of two smooth planes inclined at 40 degrees and 20 degrees to the horizontal respectively, with P on the 40-degree plane and Q on the 20-degree plane. The system is released from rest with the string taut, and P moves down its plane. (a) Show that the acceleration of the system is given by a = g(5sin40 - 2sin20)/7. (4) (b) Find the tension in the string, giving your answer to 3 significant figures. (3)
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A particle of mass 2 kg lies on a rough plane inclined at 25 degrees to the horizontal. A horizontal force of magnitude 25 N, acting in the vertical plane containing the line of greatest slope, is applied to the particle, pushing it up the plane. The coefficient of friction between the particle and the plane is 0.3, and the particle is moving up the plane. Take g = 9.8 m/s^2. (a) Show that the normal reaction between the particle and the plane is R = 2g cos25 + 25 sin25 newtons, and find its value to 3 significant figures. (3) (b) Find the acceleration of the particle. (4)
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Free printable worksheet
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