Problems

Filters
Clear Filters

2 problems found

2019 Paper 2 Q10
D: 1500.0 B: 1500.0
moments friction equilibrium rigid body string tension inequality rod on rail ring on rail

A small light ring is attached to the end \(A\) of a uniform rod \(AB\) of weight \(W\) and length \(2a\). The ring can slide on a rough horizontal rail. One end of a light inextensible string of length \(2a\) is attached to the rod at \(B\) and the other end is attached to a point \(C\) on the rail so that the rod makes an angle of \(\theta\) with the rail, where \(0 < \theta < 90^{\circ}\). The rod hangs in the same vertical plane as the rail. A force of \(kW\) acts vertically downwards on the rod at \(B\) and the rod is in equilibrium.

  1. You are given that the string will break if the tension \(T\) is greater than \(W\). Show that (assuming that the ring does not slip) the string will break if $$2k + 1 > 4 \sin \theta.$$
  2. Show that (assuming that the string does not break) the ring will slip if $$2k + 1 > (2k + 3)\mu \tan \theta,$$ where \(\mu\) is the coefficient of friction between the rail and the ring.
  3. You are now given that \(\mu \tan \theta < 1\). Show that, when \(k\) is increased gradually from zero, the ring will slip before the string breaks if $$\mu < \frac{2 \cos \theta}{1 + 2 \sin \theta}.$$

Solution:

TikZ diagram
  1. \(\,\) \begin{align*} \overset{\curvearrowright}{A}:&& W \cos \theta \cdot a + kW \cos \theta \cdot 2a - T \cos \theta \sin \theta \cdot 2a - T \sin \theta \cos \theta \cdot 2a &= 0 \\ && (2k+1) \cos \theta W &= T \cos \theta \cdot 4 \sin \theta \\ \Rightarrow && T &= \frac{2k+1}{4 \sin \theta} W \\ \Rightarrow && \text{breaks if }\quad \quad 2k+1 &> 4 \sin \theta \end{align*}
  2. \(\,\) \begin{align*} \text{N2}(\uparrow): && R - W - kW - T \sin \theta &= 0 \\ \Rightarrow && R &= (k+1)W - T \sin \theta \\ &&&= (k+1)W - \frac{2k+1}{4} W \\ &&&= \frac{2k+3}{4}W \\ \text{N2}(\leftarrow): && F_A - T \cos \theta &= 0 \\ \Rightarrow && F_A &= \frac{2k+1}{4 }\cot \theta \\ \Rightarrow && \text{slips if }\quad \quad\quad \quad\quad \quad F_A &> \mu R \\ \Rightarrow && \text{slips if }\quad \quad \frac{2k+1}{4 }\cot \theta &> \mu \frac{2k+3}{4}W \\ \Rightarrow && 2k+1 &> (2k+3) \mu \tan \theta \end{align*}
  3. The condition for breaking is \(k > 2\sin \theta -\frac12\). The condition for slipping is equivalent to: \begin{align*} && 2k+1 &> (2k+3) \mu \tan \theta \\ \Leftrightarrow && 2k(1- \mu \tan \theta) &> 3 \mu \tan \theta-1 \\ \Leftrightarrow && k &> \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} \end{align*} Therefore we will slip first if: \begin{align*} && \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} &< 2 \sin \theta - \frac12 \\ \Leftrightarrow && 3 \mu \tan \theta-1 &< 4 \sin \theta (1- \mu \tan \theta) - (1- \mu \tan \theta) \\ &&&=4 \sin \theta - 1 + \mu \tan \theta (1-4 \sin \theta) \\ \Leftrightarrow && 3 \mu \tan \theta &< 4 \sin \theta + \mu \tan \theta (1- 4 \sin \theta) \\ \Leftrightarrow && \mu \tan \theta(3-1+4\sin \theta) &< 4 \sin \theta \\ \Leftrightarrow && \mu &< \frac{2 \cos \theta}{1+2 \sin \theta} \end{align*}

View
2016 Paper 1 Q9
D: 1516.0 B: 1469.4
friction rigid body moments equilibrium rod on rail limiting equilibrium trigonometric normal reaction

A horizontal rail is fixed parallel to a vertical wall and at a distance \(d\) from the wall. A uniform rod \(AB\) of length \(2a\) rests in equilibrium on the rail with the end \(A\) in contact with the wall. The rod lies in a vertical plane perpendicular to the wall. It is inclined at an angle \(\theta\) to the vertical (where \(0 < \theta < \frac12\pi\)) and \(a\sin\theta < d\), as shown in the diagram.

TikZ diagram
The coefficient of friction between the rod and the wall is \(\mu\), and the coefficient of friction between the rod and the rail is \(\lambda\). Show that in limiting equilibrium, with the rod on the point of slipping at both the wall and the rail, the angle \(\theta\) satisfies \[ d\cosec^2\theta = a\big( (\lambda+\mu)\cos\theta + (1-\lambda \mu)\sin\theta \big) \,. \] Derive the corresponding result if, instead, \( a\sin\theta > d \).

Solution:

TikZ diagram
Notice everything is at limiting equilibrium, so \(F_W = \mu R_W\) and \(F_R = \lambda R_R\). \begin{align*} \text{N2}(\nearrow): && \lambda R_R - W \cos \theta+ R_W \sin \theta+\mu R_W \cos \theta &= 0 \\ \text{N2}(\nwarrow): && R_R -W \sin \theta -R_W \cos \theta+\mu R_W \sin \theta &= 0 \\ \overset{\curvearrowright}{A}: && a W \sin \theta -R_R \frac{d}{\sin \theta} &= 0 \\ \overset{\curvearrowright}{\text{rod}}: && -W\left (d-a\sin \theta \right)+\mu R_W d-R_W d \cot \theta &= 0 \\ \end{align*} So \begin{align*} && R_W d(\mu - \cot \theta) &= W (d - a \sin \theta) \\ && a W &= R_Rd \textrm{cosec}^2 \theta \\ \Rightarrow && d \textrm{cosec}^2 \theta &=\frac{aW}{R_R} \\ && \lambda R_R &= W \cos \theta -R_W(\sin \theta + \mu \cos \theta) \\ &&&= W\cos \theta - W \frac{d - a \sin \theta}{d(\mu - \cot \theta)} ( \sin \theta + \mu \cos \theta) \\ &&&= W { \left ( \frac{d\mu \cos \theta - d\cos \theta \cot \theta - d \sin \theta - d \mu \cos \theta+a \sin^2 \theta + a \mu \sin \theta \cos \theta}{d(\mu - \cot \theta)} \right) }\\ &&&= W \left ( \frac{ - d \textrm{cosec} \theta +a \sin^2 \theta + a \mu \sin \theta \cos \theta}{d(\mu - \cot \theta)} \right) \\ \Rightarrow && d \textrm{cosec}^2 \theta &=\frac{aW}{R_R} \\ &&&= \frac{ad\lambda(\mu - \cot \theta)}{- d \textrm{cosec} \theta +a \sin^2 \theta + a \mu \sin \theta \cos \theta} \\ &&&= \frac{ad\lambda(\mu \sin \theta - \cos \theta)}{-d + a \sin^2 \theta (\sin \theta + \mu \cos \theta)} \\ \Rightarrow && -d^2 \textrm{cosec}^2 \theta &+ a(\sin \theta + \mu \cos \theta) = ad\lambda(\mu \sin \theta - \cos \theta) \\ \Rightarrow && d \textrm{cosec}^2 \theta &= a(\sin \theta + \mu \cos \theta)-a\lambda(\mu \sin \theta - \cos \theta) \\ &&&= a( (\mu+\lambda)\cos \theta + (1-\mu \lambda)\sin \theta) \end{align*} If the rod is before the midpoint, the directions of both frictions will be reversed, ie we should obtain the same result, but with \(\mu \to -\mu, \lambda \to -\lambda\) ie \(d \textrm{cosec}^2 \theta = a( -(\mu+\lambda)\cos \theta + (1-\mu \lambda)\sin \theta)\)

View