Standard +0.8 This is a non-trivial equilibrium problem requiring resolution of forces in two directions (parallel and perpendicular to the plane), understanding that friction can act in either direction depending on P, and finding the critical value where friction reverses. It goes beyond standard textbook exercises by requiring students to identify that maximum P occurs when friction acts down the slope, not just applying F=μR mechanically.
4 A particle of mass 12 kg is stationary on a rough plane inclined at an angle of \(25 ^ { \circ }\) to the horizontal. A pulling force of magnitude \(P \mathrm {~N}\) acts at an angle of \(8 ^ { \circ }\) above a line of greatest slope of the plane. This force is used to keep the particle in equilibrium. The coefficient of friction between the particle and the plane is 0.3 .
Find the greatest possible value of \(P\).
For attempting to solve for \(P\), using equations with the correct number of relevant terms in both.
\(P = 80.8\)
A1
From \(P = 80.755...\) Allow \(P \leqslant 80.8\). If more than one case is considered for direction of friction then a choice must be made for final answer.
Alternative mark scheme for Question 4:
Answer
Marks
Guidance
Answer
Marks
Guidance
For resolving forces either vertically or horizontally
M1
Correct number of terms in either equation.
\(R\cos 25 + P\sin 33 = 12g + F\sin 25\)
A1
\(P\cos 33 = F\cos 25 + R\sin 25\)
A1
\(F = 0.3R\)
M1
Use \(F = 0.3R\)
Solve a pair of simultaneous equations in \(P\) and \(R\). May see \(R = 97.5\)
M1
For attempting to solve for \(P\), using equations with the correct number of relevant terms.
\(P = 80.8\)
A1
From \(P = 80.755...\) Allow \(P \leqslant 80.8\). If more than one case is considered for direction of friction then a choice must be made for final answer.
6
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| For resolving either parallel to or perpendicular to the plane | M1 | Three relevant terms in either equation. |
| $P\cos 8 = F + 12g\sin 25$ | A1 | |
| $12g\cos 25 = R + P\sin 8$ | A1 | |
| $F = 0.3R$ | M1 | Use $F = 0.3R$, where $R$ must involve components of both $12g$ and $P$. |
| $P\cos 8 = 0.3(12g\cos 25 - P\sin 8) + 12g\sin 25$ | M1 | For attempting to solve for $P$, using equations with the correct number of relevant terms in both. |
| $P = 80.8$ | A1 | From $P = 80.755...$ Allow $P \leqslant 80.8$. If more than one case is considered for direction of friction then a choice must be made for final answer. |
**Alternative mark scheme for Question 4:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| For resolving forces either vertically or horizontally | M1 | Correct number of terms in either equation. |
| $R\cos 25 + P\sin 33 = 12g + F\sin 25$ | A1 | |
| $P\cos 33 = F\cos 25 + R\sin 25$ | A1 | |
| $F = 0.3R$ | M1 | Use $F = 0.3R$ |
| Solve a pair of simultaneous equations in $P$ and $R$. May see $R = 97.5$ | M1 | For attempting to solve for $P$, using equations with the correct number of relevant terms. |
| $P = 80.8$ | A1 | From $P = 80.755...$ Allow $P \leqslant 80.8$. If more than one case is considered for direction of friction then a choice must be made for final answer. |
| | **6** | |
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4 A particle of mass 12 kg is stationary on a rough plane inclined at an angle of $25 ^ { \circ }$ to the horizontal. A pulling force of magnitude $P \mathrm {~N}$ acts at an angle of $8 ^ { \circ }$ above a line of greatest slope of the plane. This force is used to keep the particle in equilibrium. The coefficient of friction between the particle and the plane is 0.3 .
Find the greatest possible value of $P$.\\
\hfill \mbox{\textit{CAIE M1 2021 Q4 [6]}}