| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod with end on ground or wall supported by string |
| Difficulty | Standard +0.3 This is a standard M2 equilibrium problem requiring resolution of forces and taking moments about a point. While it involves a non-uniform rod and limiting equilibrium with friction, the solution follows a routine method: resolve vertically and horizontally, take moments, apply F=μR, and solve algebraically. The trigonometry is straightforward with tan θ = 5/12 given. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
5.
A non-uniform rod $A B$, of mass 5 kg and length 4 m , rests with one end $A$ on rough horizontal ground. The centre of mass of the rod is $d$ metres from $A$. The rod is held in limiting equilibrium at an angle $\theta$ to the horizontal by a force $\mathbf { P }$, which acts in a direction perpendicular to the rod at $B$, as shown in Figure 2. The line of action of $\mathbf { P }$ lies in the same vertical plane as the rod.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $d , g$ and $\theta$,
\begin{enumerate}[label=(\roman*)]
\item the magnitude of the vertical component of the force exerted on the rod by the ground,
\item the magnitude of the friction force acting on the rod at $A$.
Given that $\tan \theta = \frac { 5 } { 12 }$ and that the coefficient of friction between the rod and the ground is $\frac { 1 } { 2 }$,
\end{enumerate}\item find the value of $d$.\\
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\hfill \mbox{\textit{Edexcel M2 2016 Q5 [12]}}