Standard +0.8 This is a non-trivial equilibrium problem requiring resolution of forces in two directions on an inclined plane with a horizontal force (not parallel to slope), friction that could act either up or down the slope, and optimization to find the minimum force. Requires careful geometric reasoning about force directions and understanding when friction reverses, going beyond standard textbook slope problems.
5
\includegraphics[max width=\textwidth, alt={}, center]{e5ee28f2-5876-4149-9a77-18c5792c1bd8-07_366_567_258_790}
A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of \(35 ^ { \circ }\) to the horizontal. The particle is kept in equilibrium by a horizontal force of magnitude \(P \mathrm {~N}\) acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and plane is 0.4 .
Find the least possible value of \(P\).
Attempt at resolving parallel or perpendicular to the plane
\*M1
3 terms, allow sign errors, allow sin/cos mix, allow \(g\) missing. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved
Where \(R\) is initially a linear combination of a \(P\) component and a weight component (or mass component)
Solve for \(P\)
DM1
From equations with correct number of relevant resolved terms. \(R=\frac{0.6g}{\cos35+0.4\sin35}=5.7222\). Must get to \(P=\ldots\), e.g. \(P=\frac{0.6g\sin35-0.4\times0.6g\cos35}{\cos35+0.4\sin35}\). If no working seen, allow this mark if correct solution for their equations. If \(F\leqslant0.4R\) used, it should be used correctly, e.g. \(0.6g\sin35-P\cos35\leqslant0.4(P\sin35+0.6g\cos35)\)
\(P=1.41\)
A1
AWRT 1.41. If \(P\geqslant1.41\) seen, must then state the least value explicitly for A1
Question 5 (Alternative: Resolving vertically and horizontally):
Answer
Marks
Guidance
Answer
Mark
Guidance
Attempt at resolving vertically or horizontally
\*M1
3 terms, allow sign errors, allow sin/cos mix, allow \(g\) missing. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved.
\(R\cos 35 + F\sin 35 = 0.6g\)
A1
Their \(F\) or \(R\).
\(P + F\cos 35 = R\sin 35\)
A1
Their \(F\) or \(R\).
Use of \(F = 0.4R\)
\*M1
To get 2 equations, one in \(R\) (or \(F\)) and the other in \(P\) and \(R\) (or \(P\) and \(F\)) from resolved equations with correct number of relevant terms. Allow \(g\) missing.
Solve for \(P\)
DM1
From equations with the correct number of relevant resolved terms. May see \(R = \dfrac{0.6g}{\cos 35 + 0.4\sin 35} = 5.7222\); Must get to \(P = \ldots\), e.g. \(P = \dfrac{0.6g\sin 35 - 0.4\times 0.6g\cos 35}{\cos 35 + 0.4\sin 35}\). If no working seen, allow this mark if correct solution for *their* equations.
\(P = 1.41\)
A1
AWRT \(1.41\).
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at resolving parallel or perpendicular to the plane | \*M1 | 3 terms, allow sign errors, allow sin/cos mix, allow $g$ missing. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved |
| $R = P\sin35 + 0.6g\cos35\ [R=(0.573\ldots)P+4.914\ldots]$ | A1 | |
| $F + P\cos35 = 0.6g\sin35\ [F+(0.819\ldots)P=3.441\ldots]$ | A1 | Their $F$ |
| Use of $F=0.4R$ | \*M1 | Where $R$ is initially a linear combination of a $P$ component and a weight component (or mass component) |
| Solve for $P$ | DM1 | From equations with correct number of relevant resolved terms. $R=\frac{0.6g}{\cos35+0.4\sin35}=5.7222$. Must get to $P=\ldots$, e.g. $P=\frac{0.6g\sin35-0.4\times0.6g\cos35}{\cos35+0.4\sin35}$. If no working seen, allow this mark if correct solution for their equations. If $F\leqslant0.4R$ used, it should be used correctly, e.g. $0.6g\sin35-P\cos35\leqslant0.4(P\sin35+0.6g\cos35)$ |
| $P=1.41$ | A1 | AWRT 1.41. If $P\geqslant1.41$ seen, must then state the least value explicitly for A1 |
## Question 5 (Alternative: Resolving vertically and horizontally):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt at resolving vertically or horizontally | \*M1 | 3 terms, allow sign errors, allow sin/cos mix, allow $g$ missing. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved. |
| $R\cos 35 + F\sin 35 = 0.6g$ | A1 | Their $F$ or $R$. |
| $P + F\cos 35 = R\sin 35$ | A1 | Their $F$ or $R$. |
| Use of $F = 0.4R$ | \*M1 | To get 2 equations, one in $R$ (or $F$) and the other in $P$ and $R$ (or $P$ and $F$) from resolved equations with correct number of relevant terms. Allow $g$ missing. |
| Solve for $P$ | DM1 | From equations with the correct number of relevant resolved terms. May see $R = \dfrac{0.6g}{\cos 35 + 0.4\sin 35} = 5.7222$; Must get to $P = \ldots$, e.g. $P = \dfrac{0.6g\sin 35 - 0.4\times 0.6g\cos 35}{\cos 35 + 0.4\sin 35}$. If no working seen, allow this mark if correct solution for *their* equations. |
| $P = 1.41$ | A1 | AWRT $1.41$. |
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5\\
\includegraphics[max width=\textwidth, alt={}, center]{e5ee28f2-5876-4149-9a77-18c5792c1bd8-07_366_567_258_790}
A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of $35 ^ { \circ }$ to the horizontal. The particle is kept in equilibrium by a horizontal force of magnitude $P \mathrm {~N}$ acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and plane is 0.4 .
Find the least possible value of $P$.\\
\hfill \mbox{\textit{CAIE M1 2023 Q5 [6]}}