Problems

Filters
Clear Filters

1 problem found

2013 Paper 1 Q11
D: 1500.0 B: 1500.0
friction equilibrium resolving forces angle of friction strings block on surface inequality trigonometric identity

\(\,\)

TikZ diagram
The diagram shows a small block \(C\) of weight \(W\) initially at rest on a rough horizontal surface. The coefficient of friction between the block and the surface is \(\mu\). Two light strings, \(AC\) and \(BC\), are attached to the block, making angles \(\frac12 \pi -\alpha\) and \(\alpha\) to the horizontal, respectively. The tensions in \(AC\) and \(BC\) are \(T\sin\beta\) and \(T\cos\beta\) respectively, where \(0< \alpha+\beta<\frac12\pi\).
  1. In the case \(W> T\sin(\alpha+\beta)\), show that the block will remain at rest provided \[ W\sin\lambda \ge T\cos(\alpha+\beta- \lambda)\,, \] where \(\lambda\) is the acute angle such that \(\tan\lambda = \mu\).
  2. In the case \(W=T\tan\phi\), where \(2\phi =\alpha+\beta\), show that the block will start to move in a direction that makes an angle \(\phi\) with the horizontal.

Solution:

TikZ diagram
  1. Assuming the block is at rest we must have: \begin{align*} \text{N2}(\uparrow): && 0 &= T \sin \beta\cos \alpha + T \cos \beta \sin \alpha +R -W \\ \Rightarrow && W &> T \sin \beta\cos \alpha + T \cos \beta \sin \alpha \\ &&&= T\sin(\alpha+\beta) \\ \Rightarrow && R &= W-T\sin(\alpha+\beta)\\ \\ \text{N2}(\rightarrow): && 0 &= T \cos \beta \cos \alpha - T \sin \beta \sin \alpha - F \\ \Rightarrow && T \cos(\alpha+\beta) &= F \\ &&&\leq \mu (W-T\sin(\alpha+\beta)) \\ \Rightarrow && W \sin \lambda &\geq T \cos (\alpha+\beta)\cos \lambda +T \sin (\alpha+\beta) \sin \lambda \\ &&&= T\cos(\alpha+\beta-\lambda) \end{align*}
  2. If \(W = T\tan \phi\) where \(2\phi = \alpha + \beta\) then \begin{align*} \text{N2}(\uparrow): && ma_y &= T\sin(\alpha+\beta) - W \\ &&&= T \sin(\alpha+\beta) - T \tan \left ( \frac{\alpha+\beta}{2} \right ) \\ &&&= T \tan \left ( \frac{\alpha+\beta}{2} \right ) \left ( 2 \cos^2 \left ( \frac{\alpha+\beta}{2} \right ) -1\right) \\ &&&= T \tan \phi \cos \left ( \alpha+\beta\right ) \tag{notice this is positive so \(R=F=0\)} \\ \text{N2}(\rightarrow): && ma_x &= T \cos(\alpha+\beta) \\ \Rightarrow && \frac{a_y}{a_x} &= \tan \phi \end{align*} Therefore we are accelerating at an angle \(\phi\) to the horizontal

View