CAIE M1 2014 June — Question 1 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeSingle angled force - find limiting friction or coefficient
DifficultyModerate -0.8 This is a straightforward two-part mechanics problem requiring resolution of forces and application of the limiting friction formula F = μR. Part (i) involves simple vertical equilibrium (R = 7g - X cos 15°), and part (ii) requires horizontal equilibrium with F = μR. Both are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the angled force component.
Spec3.03m Equilibrium: sum of resolved forces = 03.03u Static equilibrium: on rough surfaces

1 \includegraphics[max width=\textwidth, alt={}, center]{139371b7-e142-4ed6-bff3-3ca4c32b9c6b-2_426_424_258_863} A block \(B\) of mass 7 kg is at rest on rough horizontal ground. A force of magnitude \(X \mathrm {~N}\) acts on \(B\) at an angle of \(15 ^ { \circ }\) to the upward vertical (see diagram).
  1. Given that \(B\) is in equilibrium find, in terms of \(X\), the normal component of the force exerted on \(B\) by the ground.
  2. The coefficient of friction between \(B\) and the ground is 0.4 . Find the value of \(X\) for which \(B\) is in limiting equilibrium.

Question 1:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(N + \text{component of } X = \text{Weight of } B\)M1 For resolving forces acting on the block vertically (3 terms required)
Normal component is \((70 - X\cos15°)\) NA1 [2]
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = X\sin15°\)B1
\(X\sin15° = 0.4(70 - X\cos15°)\)M1 For using \(F = \mu R\)
Value of \(X\) is \(43.4\)A1 [3]
## Question 1:

### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N + \text{component of } X = \text{Weight of } B$ | M1 | For resolving forces acting on the block vertically (3 terms required) |
| Normal component is $(70 - X\cos15°)$ N | A1 [2] | |

### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = X\sin15°$ | B1 | |
| $X\sin15° = 0.4(70 - X\cos15°)$ | M1 | For using $F = \mu R$ |
| Value of $X$ is $43.4$ | A1 [3] | |

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1\\
\includegraphics[max width=\textwidth, alt={}, center]{139371b7-e142-4ed6-bff3-3ca4c32b9c6b-2_426_424_258_863}

A block $B$ of mass 7 kg is at rest on rough horizontal ground. A force of magnitude $X \mathrm {~N}$ acts on $B$ at an angle of $15 ^ { \circ }$ to the upward vertical (see diagram).\\
(i) Given that $B$ is in equilibrium find, in terms of $X$, the normal component of the force exerted on $B$ by the ground.\\
(ii) The coefficient of friction between $B$ and the ground is 0.4 . Find the value of $X$ for which $B$ is in limiting equilibrium.

\hfill \mbox{\textit{CAIE M1 2014 Q1 [5]}}