Edexcel M1 2006 January — Question 5 14 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2006
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeEquilibrium on slope with horizontal force
DifficultyStandard +0.3 This is a standard M1 statics problem requiring resolution of forces parallel and perpendicular to an inclined plane, with friction at limiting equilibrium. While it involves multiple steps and careful component resolution, it follows a well-established method taught in all M1 courses with no novel insight required. The three parts build systematically, making it slightly easier than average.
Spec3.03e Resolve forces: two dimensions3.03r Friction: concept and vector form3.03s Contact force components: normal and frictional3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

\includegraphics{figure_2} A parcel of weight \(10\) N lies on a rough plane inclined at an angle of \(30°\) to the horizontal. A horizontal force of magnitude \(P\) newtons acts on the parcel, as shown in Figure 2. The parcel is in equilibrium and on the point of slipping up the plane. The normal reaction of the plane on the parcel is \(18\) N. The coefficient of friction between the parcel and the plane is \(\mu\). Find
  1. the value of \(P\), [4]
  2. the value of \(\mu\). [5]
The horizontal force is removed.
  1. Determine whether or not the parcel moves. [5]

Part (a)
\(R(\text{perp to plane}): P \sin 30° + 10 \cos 30° = 18\)
AnswerMarks Guidance
Solve: \(P \approx 18.7 \text{ N}\)M1 A1 ↓ M1 A1 (4 marks)
Part (b)
\(R(\parallel \text{plane}): P \cos 30° = 10 \sin 30° + F\)
\(F = 18\mu\) used
AnswerMarks Guidance
Sub and solve: \(\mu = 0.621\) or \(0.62\)M1 ↓↓ M1 A1 (5 marks)
Part (c)
Normal reaction now \(= 10 \cos 30°\)
AnswerMarks
Component of weight down plane \(= 10 \sin 30° (= 5 \text{ N})\) (seen)M1 A1 B1 ↓ M1
\(F_{\max} = \mu R_{\text{new}} \approx 5.37 \text{ N}\) (AWRT 5.4)A1 cso
\(5.37 > 5\) ⟹ does not slide(5 marks)
Total for Question 5: 14 marks
## Part (a)
$R(\text{perp to plane}): P \sin 30° + 10 \cos 30° = 18$

Solve: $P \approx 18.7 \text{ N}$ | M1 A1 ↓ M1 A1 | (4 marks)

## Part (b)
$R(\parallel \text{plane}): P \cos 30° = 10 \sin 30° + F$

$F = 18\mu$ used

Sub and solve: $\mu = 0.621$ or $0.62$ | M1 ↓↓ M1 A1 | (5 marks)

## Part (c)
Normal reaction now $= 10 \cos 30°$

Component of weight down plane $= 10 \sin 30° (= 5 \text{ N})$ (seen) | M1 A1 B1 ↓ M1 |

$F_{\max} = \mu R_{\text{new}} \approx 5.37 \text{ N}$ (AWRT 5.4) | A1 cso |

$5.37 > 5$ ⟹ does not slide | (5 marks)

**Total for Question 5: 14 marks**

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\includegraphics{figure_2}

A parcel of weight $10$ N lies on a rough plane inclined at an angle of $30°$ to the horizontal. A horizontal force of magnitude $P$ newtons acts on the parcel, as shown in Figure 2. The parcel is in equilibrium and on the point of slipping up the plane. The normal reaction of the plane on the parcel is $18$ N. The coefficient of friction between the parcel and the plane is $\mu$. Find

\begin{enumerate}[label=(\alph*)]
\item the value of $P$, [4]
\item the value of $\mu$. [5]
\end{enumerate}

The horizontal force is removed.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{2}
\item Determine whether or not the parcel moves. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2006 Q5 [14]}}