| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| 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 mechanics question requiring basic resolution of forces into components and finding a resultant using Pythagoras and trigonometry. The steps are routine: resolve the 5N force, sum components in each direction, find magnitude and direction. No problem-solving insight needed, just direct application of standard methods taught early in M1. |
| Spec | 1.05g Exact trigonometric values: for standard angles3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5\cos30\) or \(5\sin60\) or \(4.33\) | B1 | Order immaterial, accept \(\pm\). May be awarded in (ii) if no attempt in (i) |
| \(5\cos60\) or \(5\sin30\) or \(2.5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Z^2 = 7^2 + 9^2 \, (=130, Z=11.4017...)\) | Z is resultant of 7N and 9N forces only | |
| \(\cos(\text{angle of Z with y axis}) = 9/11.4017\) | ||
| angle of Z with y axis \(= 37.8746...\) | ||
| Angle opposite R in triangle of forces \(= 180 - (37.8746 + 90 + 30)\) | M1* | R is resultant of all 3 forces. Complete method |
| \(= 22.125\) (Accept 22) | A1 | |
| \(R^2 = 5^2 + 11.4017^2 - 2 \times 5 \times 11.4017\cos22.125\) | D*M1 | Cosine rule to find R |
| \(R \, (=7.0269) = 7.03\) N | A1 | |
| \(11.4017^2 = 5^2 + 7.0269^2 - 2 \times 5 \times 7.0269\cosA\) | D*M1 | Or Sine Rule. A is angle between R and 5N forces |
| \((A = 142.33)\) | Complete method | |
| Angle between R and y axis \(= 142.33 - 90 \, (=22.33)\) | ||
| \(\theta \, (= 90 - 22.33) = 67.7\) degrees | A1 | \(\theta\) is angle between R and x axis |
# Question 3:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5\cos30$ or $5\sin60$ or $4.33$ | B1 | Order immaterial, accept $\pm$. May be awarded in (ii) if no attempt in (i) |
| $5\cos60$ or $5\sin30$ or $2.5$ | B1 | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Z^2 = 7^2 + 9^2 \, (=130, Z=11.4017...)$ | | Z is resultant of 7N and 9N forces only |
| $\cos(\text{angle of Z with y axis}) = 9/11.4017$ | | |
| angle of Z with y axis $= 37.8746...$ | | |
| Angle opposite R in triangle of forces $= 180 - (37.8746 + 90 + 30)$ | M1* | R is resultant of all 3 forces. Complete method |
| $= 22.125$ (Accept 22) | A1 | |
| $R^2 = 5^2 + 11.4017^2 - 2 \times 5 \times 11.4017\cos22.125$ | D*M1 | Cosine rule to find R |
| $R \, (=7.0269) = 7.03$ N | A1 | |
| $11.4017^2 = 5^2 + 7.0269^2 - 2 \times 5 \times 7.0269\cosA$ | D*M1 | Or Sine Rule. A is angle between R and 5N forces |
| $(A = 142.33)$ | | Complete method |
| Angle between R and y axis $= 142.33 - 90 \, (=22.33)$ | | |
| $\theta \, (= 90 - 22.33) = 67.7$ degrees | A1 | $\theta$ is angle between R and x axis |3\\
\includegraphics[max width=\textwidth, alt={}, center]{470e70de-66ba-4dcc-a205-0c92f29471b1-2_570_679_1512_731}
Three horizontal forces act at the point $O$. One force has magnitude 7 N and acts along the positive $x$-axis. The second force has magnitude 9 N and acts along the positive $y$-axis. The third force has magnitude 5 N and acts at an angle of $30 ^ { \circ }$ below the negative $x$-axis (see diagram).\\
(i) Find the magnitudes of the components of the 5 N force along the two axes.\\
(ii) Calculate the magnitude of the resultant of the three forces. Calculate also the angle the resultant makes with the positive $x$-axis.
\hfill \mbox{\textit{OCR M1 2009 Q3 [8]}}