Solve a rough incline linked to a hanging particle
Couple along-slope and vertical balances while retaining friction’s direction and testing the rest state or stopping event.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use weight components and friction on an incline.
- Apply ideal-string constraints and simultaneous equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A slope particle moves up and its string is parallel to the slope. Which friction direction enters its equation?
What you will learn
- Solve a mixed incline-and-hanging connected system.
- Check that acceleration direction and static restart agree with the stated motion.
Choose coordinates along each particle’s path
A 5 kg particle on a slope is connected over a smooth pulley to a hanging 4 kg mass. Suppose the 4 kg mass moves down while the slope particle moves up, with sin θ = 3/5, cos θ = 4/5 and μ = 1/10. Reaction on the slope is 40 N, sliding friction is 4 N down the slope and the downslope weight component is 30 N. Equations are T − 30 − 4 = 5a and 40 − T = 4a, giving a = 2/3 m s⁻² and T = 112/3 N.
The normal reaction is determined by the slope particle’s perpendicular balance, not the hanging weight or string tension when the string is parallel to the slope. Sum the along-motion equations to cancel tension and solve for a, then substitute into both individual balances. If the result a is negative on these chosen axes, the specified upward/downward motion slows rather than instantly reverses.
Reverse friction only when the motion actually reverses
If the slope particle instead moves down, its friction acts up. Choose slope-down and hanging-up positive, giving mg sin θ − μR − T = ma and T − Mg = Ma. With slope mass 5 kg, hanging mass 2 kg and the same slope and μ, total drive is 30 − 4 − 20 = 6 N, so a = 6/7 m s⁻² down the slope and tension is 152/7 N. Those equations use a different sliding direction from the first example.
For a system released from rest, compare the difference of gravitational drives with the static-friction limit before choosing a sliding branch. A 5 kg slope particle and 3 kg hanging particle on the same slope balance gravitational drives at 30 N each; static friction may be zero. With 3.2 kg hanging, rest requires friction 2 N down the plane, below the 4 N limit. At a stopping event, this equilibrium test replaces the preceding sliding equation.
Pause and explain
For T − 34 = 5a and 40 − T = 4a, which acceleration follows?
Worked example
A 5 kg slope particle moves up while a 4 kg hanging particle descends on an ideal taut string. Given sin θ = 3/5, cos θ = 4/5 and μ = 1/10, find a and T.
Show the worked solution
- Slope reaction is 40 N, friction is 4 N down and the downslope weight component is 30 N.
- Add T − 34 = 5a and 40 − T = 4a to get a = 6/9 = 2/3 m s⁻².
- T = 40 − 4(2/3) = 112/3 N; both separate balances hold.
Answer a = 2/3 m s⁻², slope up and hanging down; T = 112/3 N.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 3 kg particle moves up a smooth 30° slope while its connected 2 kg partner descends on an ideal pulley string. Find acceleration and tension.
Give me a hint
The slope’s opposing gravitational component is 30 sin 30°, with no friction.
Compare my reasoning
- Equations are T − 15 = 3a and 20 − T = 2a.
- Adding gives 5 = 5a, so a = 1 m s⁻².
- T = 15 + 3 × 1 = 18 N, checked in the hanging equation.
a = 1 m s⁻², slope up; T = 18 N.
Look for these in your work
- Use the along-slope weight component.
- Apply the common acceleration magnitude while the string is taut.
A 5 kg particle moves up a rough slope connected to a descending 4 kg hanging mass. With sin θ = 3/5, cos θ = 4/5 and μ = 1/10, find a and T.
Give me a hint
Friction and the slope weight component both oppose the upward slope motion.
Compare my reasoning
- Reaction is 40 N and friction is 4 N, so slope equation is T − 34 = 5a.
- Together with 40 − T = 4a, this gives a = 2/3 m s⁻².
- T = 112/3 N; T − 34 = 10/3 = 5a confirms the slope balance.
a = 2/3 m s⁻²; T = 112/3 N.
Look for these in your work
- Find friction using the slope particle’s reaction.
- Check each particle rather than just the summed equation.
A 5 kg slope particle and 3.2 kg hanging particle are connected and released from rest. The slope has sin θ = 3/5, cos θ = 4/5 and static μ = 1/10. Test rest and find tension and slope friction in a resting state.
Give me a hint
Compare the difference of the two gravitational drives with the static limit.
Compare my reasoning
- Slope reaction is 40 N and static-friction limit is 4 N; downslope weight is 30 N.
- For the hanging mass at rest, T = 32 N. The slope balance requires friction 32 − 30 = 2 N down the plane.
- Since 2 ≤ 4, rest is admissible with a = 0; no sliding direction needs to be invented.
Rest is possible: T = 32 N; friction 2 N down the plane; a = 0.
Look for these in your work
- Use adjustable static friction rather than its maximum.
- Check both equilibrium equations and the friction bound.
Common mistakes
- The hanging mass’s weight is not the slope reaction.
- Choose friction from the actual slope motion.
- A negative solved acceleration can describe slowing in the stated direction.
What can you explain now?
May a released slope-and-hanging system remain at rest even when the gravitational drives differ slightly?
Compare with the explanation
Yes, if static friction can supply that difference within its limit.
The static force is the needed balance, not automatically the maximum available friction.
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Make it stick
Resolve on the slope, balance the hanging mass, then test tension, friction direction and the relevant static or sliding state.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.