2 problems found
In this question take \(g = 10 ms^{-2}.\) The point \(A\) lies on a fixed rough plane inclined at \(30^{\circ}\) to the horizontal and \(\ell\) is the line of greatest slope through \(A\). A particle \(P\) is projected up \(\ell\) from \(A\) with initial speed \(6\)ms\(^{-1}\). A time \(T\) seconds later, a particle \(Q\) is projected from \(A\) up \(\ell\), also with speed \(6\)ms\(^{-1}\). The coefficient of friction between each particle and the plane is \(1/(5\sqrt{3})\,\) and the mass of each particle is \(4\)kg.
Solution: Since the particles are identical and are projected with the same speed, the only way they can reach the same point \(x\) at the same time, is if \(A\) has reached it's apex and started descending. Considering \(P\), we must have (setting the level of \(A\) to be the \(0\) G.P.E. level), suppose it travels a distance \(x\) before becoming stationary: \begin{align*} \text{N2}(\nwarrow): && R - 4g \cos(30^\circ) &= 0 \\ \Rightarrow && R &= 20\sqrt{3} \\ \Rightarrow && \mu R &= \frac1{5 \sqrt{3}} (20 \sqrt{3}) \\ &&&= 4 \\ \end{align*} Therefore in the two phases of the journey the particle is being accelerated down the slope by either \(6\) or \(4\). \(v^2 = u^2 + 2as \Rightarrow 0 = 36 - 12s \Rightarrow s = 3\). \(v = u + at \Rightarrow t = 1\). Therefore after \(1\) second \(P\) reaches its highest point having travelled \(3\) metres. It will pass back to the start in \(s = ut + \frac12 a t^2 \Rightarrow 3 = \frac12 4 t^2 \Rightarrow t = \sqrt{3/2}\) seconds, ie the constraint is that the particle hasn't already past \(Q\) before the collision. The collision will occur when \(s = 6t - \frac12 6 t^2\) and \(s =3 - \frac12 4 (t+T-1)^2\) coincide, ie: \begin{align*} && 6t - 3t^2 &= 3 - 2(t+T-1)^2 \\ && 0 &= 3 -2(T-1)^2 -(4(T-1)+6)t + t^2 \\ && 0 &= 3 -2(T-1)^2 -(4T+2)t + t^2 \\ \Rightarrow && t &= \frac{4T+2 \pm \sqrt{(4T+2)^2 - 4(3-2(T-1)^2)}}{2} \\ &&&= \frac{4T+2 \pm \sqrt{24T^2}}{2} \\ &&&= 2T + 1 \pm \sqrt{6} T \\ &&&= (2 \pm \sqrt{6})T + 1 \end{align*} we must take the smaller root, ie \((2-\sqrt{6})T + 1\). In the case the collision occurs exactly at the start, the particle \(P\) has traveled \(6\) meters, against a force of \(4\) newtons of friction, ie work done is \(24\) Joules.
A uniform ladder of mass \(M\) rests with its upper end against a smooth vertical wall, and with its lower end on a rough slope which rises upwards towards the wall and makes an angle of \(\phi\) with the horizontal. The acute angle between the ladder and the wall is \(\theta\). If the ladder is in equilibrium, show that \(N\) and \(F\), the normal reaction and frictional force at the foot of the ladder are given by \[ N=Mg\left(\cos\phi-\frac{\tan\theta\sin\phi}{2}\right), \] \[ F=Mg\left(\sin\phi+\frac{\tan\theta\cos\phi}{2}\right). \] If the coefficient of friction between the ladder and the slope is \(2\), and \(\phi=45^{\circ}\), what is the largest value of \(\theta\) for which the ladder can rest in equilibrium?
Solution: \begin{align*} \overset{\curvearrowleft}{X}: && 0&= \frac{l}{2} Mg\sin \theta - l R_1 \cos \theta \\ \Rightarrow && R_1 &= \frac12 \tan \theta Mg \\ \text{N2}(\uparrow): && 0 &= R\cos \phi +F \sin \phi - Mg \\ \text{N2}(\rightarrow):&& 0&=R_1-F \cos \phi + R \sin \phi \\ \Rightarrow && \frac12 \tan \theta Mg &= F \cos \phi- R \sin \phi \\ && Mg &= F \sin \phi +R \cos \phi \\ \Rightarrow && F &= Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) \\ && N &= Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \end{align*} If \(\mu = 2\) and \(\phi = 45^{\circ}\), we must have \(F \leq 2 N\), so: \begin{align*} && Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) &\leq 2 Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \\ \Rightarrow && 1 + \frac12 \tan \theta \leq 2-\tan \theta \\ \Rightarrow && \frac 32 \tan \theta \leq 1 \\ \Rightarrow && \tan \theta \leq \frac23 \\ \Rightarrow && \theta \leq \tan^{-1} \frac23 \end{align*}