OCR M1 — Question 2 8 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeResultant of coplanar forces
DifficultyModerate -0.8 This is a straightforward M1 resolving forces question with given trigonometric values. Students simply resolve each force into components using the provided sin/cos values, sum them, and find the resultant magnitude using Pythagoras. No problem-solving insight required—pure mechanical application of standard techniques with arithmetic that works out cleanly (y-component = 0 by design).
Spec3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors

\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 \(\cos \beta = 0.6\).
  1. Show that the component, in the \(y\)-direction, of the resultant of the three forces is zero. [4]
  2. Find the magnitude of the resultant of the three forces. [3]
  3. State the direction of the resultant of the three forces. [1]

\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  Q2 [8]}}