Standard +0.8 This is a non-trivial mechanics problem requiring resolution of a horizontal force in two directions on an inclined plane, combined with limiting friction. Students must handle tan α = 2.4 (requiring sin/cos calculation), resolve the horizontal force P both parallel and perpendicular to the slope, apply F = μR, and solve simultaneous equations. The horizontal force (rather than parallel to slope) adds conceptual complexity beyond standard slope problems.
4
\includegraphics[max width=\textwidth, alt={}, center]{79b90ef5-ef3a-4c59-b662-d0fbfba813ca-2_365_493_1749_826}
A rough plane is inclined at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = 2.4\). A small block of mass 0.6 kg is held at rest on the plane by a horizontal force of magnitude \(P \mathrm {~N}\). This force acts in a vertical plane through a line of greatest slope (see diagram). The coefficient of friction between the block and the plane is 0.4 . The block is on the point of slipping down the plane. By resolving forces parallel to and perpendicular to the inclined plane, or otherwise, find the value of \(P\). [0pt]
[8]
For finding \(R\) and substituting into an expression for \(P\)
\(P = 6.12\)
A1
8 marks total
## Question 4:
**Main Scheme:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (resolving parallel to plane) | M1 | For resolving three forces parallel to the plane |
| $0.6g\sin\alpha = F + P\cos\alpha$ | A1 | Value of $\alpha$ used or values of $\sin\alpha$ and $\cos\alpha$ used |
| (resolving perpendicular to plane) | M1 | For resolving three forces perpendicular to the plane |
| $R = 0.6g\cos\alpha + P\sin\alpha$ | A1 | Value of $\alpha$ used or values of $\sin\alpha$ and $\cos\alpha$ used |
| (using $F = \mu R$) | M1 | For using $F = \mu R$ |
| $0.6g\sin\alpha - P\cos\alpha = 0.4(0.6g\cos\alpha + P\sin\alpha)$ | A1 | Value of $\alpha$ used or values of $\sin\alpha$ and $\cos\alpha$ used |
| $6(12/13) - P(5/13) = 2.4(5/13) + 0.4P(12/13)$ | M1 | For solving the resultant equation for $P$ |
| $P = 6.12$ | A1 | **8 marks total** |
**Alternative Scheme:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (resolving vertically) | M1 | For resolving three forces vertically |
| $W = R\cos\alpha + F\sin\alpha$ | A1 | Value of $\alpha$ used or values of $\sin\alpha$ and $\cos\alpha$ used |
| (resolving horizontally) | M1 | For resolving three forces horizontally |
| $P = R\sin\alpha - F\cos\alpha$ | A1 | Value of $\alpha$ used or values of $\sin\alpha$ and $\cos\alpha$ used |
| (using $F = \mu R$ in both equations) | M1 | For using $F = \mu R$ in both equations |
| $0.6g = R(5/13) + 0.4R(12/13)$ and $P = R(12/13) - 0.4R(5/13)$ | A1 | Value of $\alpha$ used or values of $\sin\alpha$ and $\cos\alpha$ used |
| $78 = R(5+4.8)$ and $13P = R(12-2) \rightarrow 13P = (78 \div 9.8) \times 10$ | M1 | For finding $R$ and substituting into an expression for $P$ |
| $P = 6.12$ | A1 | **8 marks total** |
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\includegraphics[max width=\textwidth, alt={}, center]{79b90ef5-ef3a-4c59-b662-d0fbfba813ca-2_365_493_1749_826}
A rough plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = 2.4$. A small block of mass 0.6 kg is held at rest on the plane by a horizontal force of magnitude $P \mathrm {~N}$. This force acts in a vertical plane through a line of greatest slope (see diagram). The coefficient of friction between the block and the plane is 0.4 . The block is on the point of slipping down the plane. By resolving forces parallel to and perpendicular to the inclined plane, or otherwise, find the value of $P$.\\[0pt]
[8]
\hfill \mbox{\textit{CAIE M1 2013 Q4 [8]}}