| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward M1 resolving forces question with given trigonometric values, requiring only standard component resolution and Pythagoras. Part (i) is a 'show that' with provided trig values eliminating calculation difficulty, part (ii) is direct application of Pythagoras, and part (iii) requires only stating a direction. No problem-solving insight needed, just methodical application of basic mechanics techniques. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(15 \times 0.28\) and \(11x 0.8\) \(Y= 15x0.28 + 11x0.8 - 13\) Component is zero AG | M1 A1 A1ft A1 | For resolving forces vertically Allow use of \(\theta = 16.3\) and \(\theta=53.1\) Ft cv(15 × 0.28 and 11x 0.8) |
| (ii) \(X = 15 \times 0.96 - 11 \times 0.6\) Magnitude is 7.8N | M1 A1 A1 | For resolving forces horizontally Allow use of \(\theta = 16.3\) and \(\theta=53.1\) Accept 7.79, -7.8 |
| (iii) Direction is that of the \((+ve)\) x -axis | B1 | Do not allow horizontal, 90' from vertical. Do not award if \(\theta = 16.3\) and \(\theta=53.1\) have been used. |
(i) $15 \times 0.28$ and $11x 0.8$ $Y= 15x0.28 + 11x0.8 - 13$ Component is zero AG | M1 A1 A1ft A1 | For resolving forces vertically Allow use of $\theta = 16.3$ and $\theta=53.1$ Ft cv(15 × 0.28 and 11x 0.8)
(ii) $X = 15 \times 0.96 - 11 \times 0.6$ Magnitude is 7.8N | M1 A1 A1 | For resolving forces horizontally Allow use of $\theta = 16.3$ and $\theta=53.1$ Accept 7.79, -7.8
(iii) Direction is that of the $(+ve)$ x -axis | B1 | Do not allow horizontal, 90' from vertical. Do not award if $\theta = 16.3$ and $\theta=53.1$ have been used.
---\includegraphics{figure_2}
Three horizontal forces of magnitudes 15 N, 11 N and 13 N act on a particle $P$ in the directions shown in the diagram. The angles $\alpha$ and $\beta$ are such that $\sin \alpha = 0.28$, $\cos \alpha = 0.96$, $\sin \beta = 0.8$ and $\cos \beta = 0.6$.
\begin{enumerate}[label=(\roman*)]
\item Show that the component, in the $y$-direction, of the resultant of the three forces is zero. [4]
\item Find the magnitude of the resultant of the three forces. [3]
\item State the direction of the resultant of the three forces. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2007 Q2 [8]}}