OCR M1 2007 January — Question 3 8 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Year2007
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeTwo particles over single pulley
DifficultyModerate -0.3 This is a standard M1 connected particles problem with friction in limiting equilibrium. Part (i) requires straightforward application of equilibrium conditions (T = 0.3g, friction = μR = T) to find μ. Part (ii) adds a horizontal force but uses the same principles. The problem is slightly easier than average because it's a textbook setup with clear equilibrium conditions and no complex geometry or novel reasoning required.
Spec3.03k Connected particles: pulleys and equilibrium3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces

\includegraphics{figure_3} A block \(B\) of mass 0.4 kg and a particle \(P\) of mass 0.3 kg are connected by a light inextensible string. The string passes over a smooth pulley at the edge of a rough horizontal table. \(B\) is in contact with the table and the part of the string between \(B\) and the pulley is horizontal. \(P\) hangs freely below the pulley (see diagram).
  1. The system is in limiting equilibrium with the string taut and \(P\) on the point of moving downwards. Find the coefficient of friction between \(B\) and the table. [5]
  2. A horizontal force of magnitude \(X\) N, acting directly away from the pulley, is now applied to \(B\). The system is again in limiting equilibrium with the string taut, and with \(P\) now on the point of moving upwards. Find the value of \(X\). [3]

AnswerMarks Guidance
(i) \(T = 0.3g\) \(F = T\) \(R = 0.4g\) Coefficient is 0.75B1 B1 B1 M1 A1 At particle (or 0.3g -T= 0.3a) Or F = cv(T at particle) (or T - F = 0.4a) For using F = \(\mu\) R
(ii) \(X = 0.3g + 0.3g\) \(X = 5.88N\)M1 A1ft A1 For resolving 3 relevant forces on B horizontally, a=0 Ft cv(\(\mu\)) cv(R)
(i) $T = 0.3g$ $F = T$ $R = 0.4g$ Coefficient is 0.75 | B1 B1 B1 M1 A1 | At particle (or 0.3g -T= 0.3a) Or F = cv(T at particle) (or T - F = 0.4a) For using F = $\mu$ R

(ii) $X = 0.3g + 0.3g$ $X = 5.88N$ | M1 A1ft A1 | For resolving 3 relevant forces on B horizontally, a=0 Ft cv($\mu$) cv(R)

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\includegraphics{figure_3}

A block $B$ of mass 0.4 kg and a particle $P$ of mass 0.3 kg are connected by a light inextensible string. The string passes over a smooth pulley at the edge of a rough horizontal table. $B$ is in contact with the table and the part of the string between $B$ and the pulley is horizontal. $P$ hangs freely below the pulley (see diagram).

\begin{enumerate}[label=(\roman*)]
\item The system is in limiting equilibrium with the string taut and $P$ on the point of moving downwards. Find the coefficient of friction between $B$ and the table. [5]
\item A horizontal force of magnitude $X$ N, acting directly away from the pulley, is now applied to $B$. The system is again in limiting equilibrium with the string taut, and with $P$ now on the point of moving upwards. Find the value of $X$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR M1 2007 Q3 [8]}}