| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of three coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question involving perpendicular forces. Part (i) requires basic Pythagoras and inverse tan to find resultant magnitude and direction. Part (ii) simply requires understanding that equilibrium means the third force is equal and opposite to the resultant. No complex problem-solving or novel insight needed—purely routine application of standard techniques. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03p Resultant forces: using vectors |
3 Two horizontal forces $\mathbf { X }$ and $\mathbf { Y }$ act at a point $O$ and are at right angles to each other. $\mathbf { X }$ has magnitude 12 N and acts along a bearing of $090 ^ { \circ } . \mathbf { Y }$ has magnitude 15 N and acts along a bearing of $000 ^ { \circ }$.\\
(i) Calculate the magnitude and bearing of the resultant of $\mathbf { X }$ and $\mathbf { Y }$.\\
(ii) A third force $\mathbf { E }$ is now applied at $O$. The three forces $\mathbf { X } , \mathbf { Y }$ and $\mathbf { E }$ are in equilibrium. State the magnitude of $\mathbf { E }$, and give the bearing along which it acts.
\hfill \mbox{\textit{OCR M1 2008 Q3 [8]}}