| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle with string at angle to wall |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. While it involves multiple forces at angles, the method is routine and well-practiced, making it slightly easier than average for A-level. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vertically: \(T\cos40 + F\cos60 = 5\) | M1 | First equation for resolution of forces. No missing/additional terms. Condone sin/cos confusion and sign error(s). \(5g\) in place of 5 is accuracy error. \(T\) must link with 40 or 50 and \(F\) with 60 or 30 |
| Correct equation | A1 | |
| Horizontally: \(T\cos50 = F\cos30\) | M1 | Second equation for resolution of forces. No missing/additional terms. Condone sin/cos confusion and sign error(s). \(T\) must link with 40 or 50 and \(F\) with 60 or 30 |
| Correct equation | A1 | |
| Perpendicular to line of \(F\): \(T\cos10 = 5\cos30\) | ||
| Perpendicular to line of \(T\): \(F\cos10 = 5\cos50\) | ||
| Solve for \(T\) or \(F\) | dM1 | Dependent on using equation(s) that scored M mark(s) |
| \(T = 4.3969...\text{ N} = 4.4\text{ N}\) (or better) | A1 | One correct |
| \(F = 3.263.... = 3.3\text{ N}\) (or better) | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{5}{\sin100} = \dfrac{F}{\sin140} = \dfrac{T}{\sin120}\) | M1 | One pair including \(\dfrac{5}{\sin100}\) or \(\dfrac{5}{\sin80}\). Incorrect pairing of forces and angles is M0 |
| Two fractions correct | A1 | |
| Second pair of fractions | M1 | |
| All correct | A1 | |
| Solve for \(T\) or \(F\) | dM1 | Dependent on using equation(s) that scored M mark(s) |
| \(T = 4.3969...\text{ N} = 4.4\text{ N}\) (or better) | A1 | One correct |
| \(F = 3.263.... = 3.3\text{ N}\) (or better) | A1 | Both correct |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertically: $T\cos40 + F\cos60 = 5$ | M1 | First equation for resolution of forces. No missing/additional terms. Condone sin/cos confusion and sign error(s). $5g$ in place of 5 is accuracy error. $T$ must link with 40 or 50 and $F$ with 60 or 30 |
| Correct equation | A1 | |
| Horizontally: $T\cos50 = F\cos30$ | M1 | Second equation for resolution of forces. No missing/additional terms. Condone sin/cos confusion and sign error(s). $T$ must link with 40 or 50 and $F$ with 60 or 30 |
| Correct equation | A1 | |
| Perpendicular to line of $F$: $T\cos10 = 5\cos30$ | | |
| Perpendicular to line of $T$: $F\cos10 = 5\cos50$ | | |
| Solve for $T$ or $F$ | dM1 | Dependent on using equation(s) that scored M mark(s) |
| $T = 4.3969...\text{ N} = 4.4\text{ N}$ (or better) | A1 | One correct |
| $F = 3.263.... = 3.3\text{ N}$ (or better) | A1 | Both correct |
**1 alt (Lami's theorem):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{5}{\sin100} = \dfrac{F}{\sin140} = \dfrac{T}{\sin120}$ | M1 | One pair including $\dfrac{5}{\sin100}$ or $\dfrac{5}{\sin80}$. Incorrect pairing of forces and angles is M0 |
| Two fractions correct | A1 | |
| Second pair of fractions | M1 | |
| All correct | A1 | |
| Solve for $T$ or $F$ | dM1 | Dependent on using equation(s) that scored M mark(s) |
| $T = 4.3969...\text{ N} = 4.4\text{ N}$ (or better) | A1 | One correct |
| $F = 3.263.... = 3.3\text{ N}$ (or better) | A1 | Both correct |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c3869c7-008f-4131-b68d-8ecdd4da3377-02_346_499_251_721}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of weight 5 N is attached to one end of a light string. The other end of the string is attached to a fixed point $O$. A force of magnitude $F$ newtons is applied to $P$. The line of action of the force is inclined to the horizontal at $30 ^ { \circ }$ and lies in the same vertical plane as the string. The particle $P$ is in equilibrium with the string making an angle of $40 ^ { \circ }$ with the downward vertical, as shown in Figure 1.
Find\\
(i) the tension in the string,\\
(ii) the value of $F$.\\
\hfill \mbox{\textit{Edexcel M1 2017 Q1 [7]}}