10 exam-style questions on particles connected by strings, pulleys, Atwood machines and systems with friction. Tap Show answer for full working. Take \(g=10\,\mathrm{m\,s}^{-2}\) unless stated.
Two particles of masses \(3\,\mathrm{kg}\) and \(5\,\mathrm{kg}\) are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest. Find the acceleration of the system and the tension in the string.
For \(5\,\mathrm{kg}\): \(50-T=5a\). For \(3\,\mathrm{kg}\): \(T-30=3a\). Add: \(20=8a \Rightarrow a=2.5\,\mathrm{m\,s}^{-2}\). \(T=30+3\cdot2.5=37.5\,\mathrm{N}\).
A particle of mass \(4\,\mathrm{kg}\) on a smooth horizontal table is connected by a light inextensible string to a particle of mass \(2\,\mathrm{kg}\) which hangs freely. The string passes over a smooth pulley at the edge of the table. Find the acceleration of the system when released from rest with the string taut.
For hanging mass: \(20-T=2a\). For table mass: \(T=4a\). Add: \(20=6a \Rightarrow a=\dfrac{10}{3}\,\mathrm{m\,s}^{-2}\).
In an Atwood machine the two masses are \(m\) and \(3m\). Find the acceleration of the system and the tension in the string in terms of \(m\) and \(g\).
For \(3m\): \(3mg-T=3ma\). For \(m\): \(T-mg=ma\). Add: \(2mg=4ma \Rightarrow a=\dfrac{g}{2}\). \(T=mg+m\cdot\dfrac{g}{2}=\dfrac{3}{2}mg\).
Two particles \(A\) and \(B\) of masses \(2\,\mathrm{kg}\) and \(3\,\mathrm{kg}\) are connected by a light inextensible string. \(A\) is on a rough horizontal table with coefficient of friction \(0.2\); the string passes over a smooth pulley and \(B\) hangs freely. The system is released from rest. Find the acceleration and the tension. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Friction on \(A\): \(\mu R=0.2\cdot20=4\,\mathrm{N}\).
For \(B\): \(30-T=3a\). For \(A\): \(T-4=2a\). Add: \(26=5a \Rightarrow a=5.2\,\mathrm{m\,s}^{-2}\). \(T=4+2\cdot5.2=14.4\,\mathrm{N}\).
Particles \(A\) and \(B\) of masses \(4\,\mathrm{kg}\) and \(6\,\mathrm{kg}\) are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut. After \(2\,\mathrm{s}\) the string breaks. Find how much further \(B\) falls before coming to instantaneous rest. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Acceleration while connected: \(a=\dfrac{(6-4)g}{10}=2\,\mathrm{m\,s}^{-2}\). After \(2\,\mathrm{s}\), speed of \(B\) \(=4\,\mathrm{m\,s}^{-1}\) downward.
After string breaks, \(B\) is a free particle projected upward? No: \(B\) was descending, so continues downward under gravity with initial speed \(4\,\mathrm{m\,s}^{-1}\). It does not come to rest; it accelerates downward. (If the question means the lighter mass, or if \(B\) was the ascending mass, adjust. Assuming \(B\) is the heavier descending mass, it never stops. If we take the ascending mass \(A\): speed \(4\,\mathrm{m\,s}^{-1}\) upward, then \(0=16-20s \Rightarrow s=0.8\,\mathrm{m}\) further.)
A particle of mass \(m\) on a smooth plane inclined at \(30^\circ\) to the horizontal is connected by a light inextensible string which passes over a smooth pulley at the top of the plane to a particle of mass \(2m\) hanging freely. Find the acceleration of the system when released.
Assume \(2m\) descends. For \(2m\): \(2mg-T=2ma\). For \(m\): \(T-mg\sin30^\circ=ma \Rightarrow T-\dfrac{1}{2}mg=ma\).
Add: \(2mg-\dfrac{1}{2}mg=3ma \Rightarrow \dfrac{3}{2}mg=3ma \Rightarrow a=\dfrac{g}{2}\).
Two particles of masses \(m\) and \(2m\) are connected by a light inextensible string of length \(4a\). The particle of mass \(2m\) is held on a smooth horizontal table at a distance \(a\) from the edge; the string is taut and the particle of mass \(m\) hangs freely. The system is released from rest. Find the speed of each particle when the heavier particle reaches the edge of the table.
Acceleration \(a=\dfrac{mg}{3m}=\dfrac{g}{3}\). Distance travelled by each \(=a\) (until \(2m\) reaches edge). \(v^2=u^2+2as=0+2\cdot\dfrac{g}{3}\cdot a=\dfrac{2ga}{3}\). \(v=\sqrt{\dfrac{2ga}{3}}\).
Particles \(A\), \(B\) and \(C\) of masses \(2\,\mathrm{kg}\), \(3\,\mathrm{kg}\) and \(5\,\mathrm{kg}\) respectively are connected by light inextensible strings as follows: \(A\) and \(B\) are connected and rest on a smooth horizontal table; the string from \(B\) passes over a smooth pulley and is attached to \(C\) which hangs freely. The system is released from rest. Find the acceleration of the system and the tensions in the two strings.
Let tension between \(A\) and \(B\) be \(T_1\), between \(B\) and \(C\) be \(T_2\). Acceleration \(a\).
For \(C\): \(50-T_2=5a\). For \(B\): \(T_2-T_1=3a\). For \(A\): \(T_1=2a\).
From \(A\) and \(B\): \(T_2=5a\). Then \(50-5a=5a \Rightarrow 50=10a \Rightarrow a=5\,\mathrm{m\,s}^{-2}\). \(T_1=10\,\mathrm{N}\), \(T_2=25\,\mathrm{N}\).
A particle of mass \(3\,\mathrm{kg}\) on a rough horizontal table (\(\mu=0.25\)) is connected by a light inextensible string which passes over a smooth pulley to a particle of mass \(5\,\mathrm{kg}\) hanging freely. The system is released from rest. After the \(5\,\mathrm{kg}\) mass has descended \(2\,\mathrm{m}\) the string breaks. Find the total distance travelled by the \(3\,\mathrm{kg}\) mass from the start until it comes to rest. Take \(g=10\,\mathrm{m\,s}^{-2}\).
While connected: friction \(=0.25\cdot30=7.5\,\mathrm{N}\). \(50-T=5a\), \(T-7.5=3a\). Add: \(42.5=8a \Rightarrow a=5.3125\,\mathrm{m\,s}^{-2}\).
After \(2\,\mathrm{m}\): \(v^2=2\cdot5.3125\cdot2=21.25 \Rightarrow v\approx4.61\,\mathrm{m\,s}^{-1}\).
After string breaks, only friction acts on the \(3\,\mathrm{kg}\) mass: deceleration \(=\dfrac{7.5}{3}=2.5\,\mathrm{m\,s}^{-2}\). Distance to stop: \(0=21.25-2\cdot2.5\cdot s \Rightarrow s=4.25\,\mathrm{m}\).
Total distance \(=2+4.25=6.25\,\mathrm{m}\).
Two particles of equal mass \(m\) are connected by a light inextensible string of length \(2l\). One particle is held at rest on a smooth horizontal table at a distance \(l\) from a smooth hole in the table; the string passes through the hole and the other particle hangs at rest. The particle on the table is projected horizontally with speed \(u\) perpendicular to the string. Show that the hanging particle will not move if \(u^2\le gl\), and find the tension in the string in this case.
If the hanging particle remains at rest, the particle on the table moves in a circle of radius \(l\). Centripetal force \(=\dfrac{mu^2}{l}\) provided by the tension \(T\).
For the hanging particle to remain at rest, \(T\le mg\). Thus \(\dfrac{mu^2}{l}\le mg \Rightarrow u^2\le gl\).
When \(u^2\le gl\), \(T=\dfrac{mu^2}{l}\).