The diagram shows two particles \(A\) and \(B\) of masses \(\quantity{0.2}{kg}\) and \(\quantity{0.4}{kg}\) respectively lying on a rough slope inclined at an angle \(\theta\) to the horizontal where \(\tan\theta=\frac{3}{4}\). The coefficient of friction between the particles and the slope is \(\frac{1}{4}\). A light inextensible string joins \(A\) to \(B\) and the system is just maintained in equilibrium, with both particles about to move up the slope, by a force on \(B\) of \(\quantity{P\,}{N}\), acting up the slope.
[Take \(g\) to be \(\quantity{10}{m\,s^{-2}}\).]
Calculate
- the value of \(P\),
Firstly, note that the particles are about to slide up the slope so the friction force on each one will be acting down the slope and its magnitude in each case will be given by \[F=\mu R\] where \(\mu\) is the coefficient of friction, \(\frac{1}{4}\), and \(R\) is the normal reaction force. We mark up all the forces acting on the particles on a copy of the diagram.
\(T\) is the tension in the string, \(R_A\) is the normal reaction force on \(A\), \(F_A\) is the friction force on \(A\) and correspondingly for \(B\). Next, we resolve the weights into components parallel and perpendicular to the slope. We are told that \(\tan\theta=\frac{3}{4}\) and Pythagoras tells us that \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\).
The system is in equilibrium so the sum of the forces perpendicular to the slope must be zero for each particle. Using \(g=10\), we have \[R_A = 0.2g\cos\theta = 2\times\frac{4}{5}\] and \[R_B = 0.4g\cos\theta = 4\times\frac{4}{5}.\] We can now use these to calculate the magnitudes of the friction forces, \[F_A=\mu R_A = \frac{1}{4}\times2\times\frac{4}{5} = \frac{2}{5}\] and \[F_B=\mu R_B = \frac{1}{4}\times4\times\frac{4}{5} = \frac{4}{5}.\]
Now we can look at the forces parallel to the slope. We consider the two particles as a single entity so that \(T\) is an internal force and can be ignored.Alternatively, we could count \(T\) as two external forces, but since it acts up the slope on \(A\) and down the slope on \(B\) they would cancel out.
Calculate
- the tension in the string.
Resolving the forces on \(B\) instead should give exactly the same result and would be a useful check.
