12 exam-style questions on resultant forces, Newton’s laws, inclined planes and equilibrium. Tap Show answer for full working.
A particle of mass \(3\,\mathrm{kg}\) is acted on by a force of \(12\,\mathrm{N}\) horizontally. Find the acceleration of the particle and the distance it travels from rest in \(4\,\mathrm{s}\).
By Newton’s second law: \(F=ma \Rightarrow 12=3a \Rightarrow a=4\,\mathrm{m\,s}^{-2}\).
\(s=\dfrac{1}{2}at^2=\dfrac{1}{2}\cdot4\cdot16=32\,\mathrm{m}\).
Two forces of magnitudes \(5\,\mathrm{N}\) and \(12\,\mathrm{N}\) act on a particle at right angles to each other. Find the magnitude and direction of the resultant.
Magnitude: \(R=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\,\mathrm{N}\).
Direction: \(\theta=\tan^{-1}(12/5)=\tan^{-1}(2.4)\approx67.4^\circ\) from the \(5\,\mathrm{N}\) force.
A box of mass \(8\,\mathrm{kg}\) rests on a rough horizontal surface. The coefficient of friction is \(0.4\). Find the minimum horizontal force needed to move the box. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Normal reaction \(R=mg=80\,\mathrm{N}\). Limiting friction \(F=\mu R=0.4\cdot80=32\,\mathrm{N}\). Minimum force \(=32\,\mathrm{N}\).
A particle of mass \(2\,\mathrm{kg}\) is in equilibrium under the action of three coplanar forces. Two of the forces are \(6\,\mathrm{N}\) due east and \(8\,\mathrm{N}\) due north. Find the magnitude and direction of the third force.
Resultant of the two known forces: \(\sqrt{6^2+8^2}=10\,\mathrm{N}\) at \(\tan^{-1}(8/6)=\tan^{-1}(4/3)\approx53.1^\circ\) north of east.
For equilibrium the third force must be equal and opposite: \(10\,\mathrm{N}\) at \(53.1^\circ\) south of west.
A particle of mass \(5\,\mathrm{kg}\) slides down a smooth plane inclined at \(30^\circ\) to the horizontal. Find the acceleration of the particle and the distance it travels from rest in \(3\,\mathrm{s}\). Take \(g=9.8\,\mathrm{m\,s}^{-2}\).
Component of weight down the plane: \(mg\sin30^\circ=5\cdot9.8\cdot\dfrac{1}{2}=24.5\,\mathrm{N}\).
\(a=\dfrac{24.5}{5}=4.9\,\mathrm{m\,s}^{-2}\).
\(s=\dfrac{1}{2}at^2=\dfrac{1}{2}\cdot4.9\cdot9=22.05\,\mathrm{m}\).
A block of mass \(4\,\mathrm{kg}\) is placed on a rough plane inclined at \(25^\circ\) to the horizontal. The coefficient of friction is \(0.3\). Does the block remain at rest? If not, find its acceleration down the plane. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Component down plane: \(mg\sin25^\circ=40\sin25^\circ\approx16.90\,\mathrm{N}\).
Normal reaction \(R=mg\cos25^\circ=40\cos25^\circ\approx36.25\,\mathrm{N}\). Max friction \(\mu R=0.3\cdot36.25\approx10.88\,\mathrm{N}\).
Since \(16.90>10.88\), the block slides. Net force down \(=16.90-10.88=6.02\,\mathrm{N}\). \(a=\dfrac{6.02}{4}\approx1.51\,\mathrm{m\,s}^{-2}\).
A particle of mass \(m\) is suspended by two light inextensible strings of equal length. The strings make angles of \(30^\circ\) and \(60^\circ\) with the vertical. Find the tensions in the strings in terms of \(m\) and \(g\).
Resolve horizontally: \(T_1\sin30^\circ=T_2\sin60^\circ \Rightarrow \dfrac{1}{2}T_1=\dfrac{\sqrt{3}}{2}T_2 \Rightarrow T_1=\sqrt{3}\,T_2\).
Resolve vertically: \(T_1\cos30^\circ+T_2\cos60^\circ=mg \Rightarrow T_1\cdot\dfrac{\sqrt{3}}{2}+T_2\cdot\dfrac{1}{2}=mg\).
Substitute: \(\sqrt{3}\,T_2\cdot\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}T_2=mg \Rightarrow \dfrac{3}{2}T_2+\dfrac{1}{2}T_2=mg \Rightarrow 2T_2=mg \Rightarrow T_2=\dfrac{1}{2}mg\), \(T_1=\dfrac{\sqrt{3}}{2}mg\).
A lift of mass \(800\,\mathrm{kg}\) is ascending with acceleration \(1.5\,\mathrm{m\,s}^{-2}\). A person of mass \(70\,\mathrm{kg}\) stands on the floor of the lift. Find the reaction of the floor on the person and the tension in the cable supporting the lift. Take \(g=10\,\mathrm{m\,s}^{-2}\).
For the person: \(R-mg=ma \Rightarrow R-700=70\cdot1.5 \Rightarrow R=700+105=805\,\mathrm{N}\).
For the lift + person (total mass \(870\,\mathrm{kg}\)): \(T-870g=870\cdot1.5 \Rightarrow T-8700=1305 \Rightarrow T=10005\,\mathrm{N}\).
A particle of mass \(2\,\mathrm{kg}\) is on a rough horizontal surface. The coefficient of friction is \(0.25\). A force of magnitude \(P\,\mathrm{N}\) acts on the particle at an angle of \(30^\circ\) above the horizontal.
(a) Find the least value of \(P\) for which the particle moves.
(b) When \(P=10\), find the acceleration of the particle. Take \(g=10\,\mathrm{m\,s}^{-2}\).
(a) Resolve vertically: \(R+P\sin30^\circ=mg \Rightarrow R+0.5P=20 \Rightarrow R=20-0.5P\).
Horizontally for limiting friction: \(P\cos30^\circ=\mu R \Rightarrow P\cdot\dfrac{\sqrt{3}}{2}=0.25(20-0.5P)\).
\(\dfrac{\sqrt{3}}{2}P=5-0.125P \Rightarrow P(\dfrac{\sqrt{3}}{2}+0.125)=5 \Rightarrow P(0.866+0.125)=5 \Rightarrow P\approx5.05\,\mathrm{N}\).
(b) When \(P=10\): \(R=20-5=15\). Friction \(=\mu R=3.75\,\mathrm{N}\). Net force \(=10\cdot\dfrac{\sqrt{3}}{2}-3.75\approx8.66-3.75=4.91\,\mathrm{N}\). \(a=\dfrac{4.91}{2}\approx2.46\,\mathrm{m\,s}^{-2}\).
A particle of mass \(m\) rests on a rough plane inclined at angle \(\alpha\) to the horizontal, where \(\tan\alpha=\dfrac{3}{4}\). The coefficient of friction is \(\dfrac{1}{2}\). A horizontal force of magnitude \(P\) is applied to the particle so that it is on the point of moving up the plane. Find \(P\) in terms of \(m\) and \(g\).
\(\sin\alpha=\dfrac{3}{5}\), \(\cos\alpha=\dfrac{4}{5}\).
Resolve perpendicular to plane: \(R=mg\cos\alpha+P\sin\alpha=mg\cdot\dfrac{4}{5}+P\cdot\dfrac{3}{5}\).
Parallel to plane (up the plane positive, limiting): \(P\cos\alpha-mg\sin\alpha=\mu R\).
\(P\cdot\dfrac{4}{5}-mg\cdot\dfrac{3}{5}=\dfrac{1}{2}\Bigl(mg\cdot\dfrac{4}{5}+P\cdot\dfrac{3}{5}\Bigr)\).
Multiply by \(10\): \(8P-6mg=4mg+3P \Rightarrow 5P=10mg \Rightarrow P=2mg\).
Two particles \(A\) and \(B\), of masses \(3\,\mathrm{kg}\) and \(5\,\mathrm{kg}\) respectively, are connected by a light inextensible string. Particle \(A\) is on a smooth horizontal table; the string passes over a smooth pulley at the edge of the table and \(B\) hangs freely. The system is released from rest with the string taut.
(a) Find the acceleration of the system and the tension in the string.
(b) After \(2\,\mathrm{s}\), the string breaks. How far does \(A\) travel before coming to rest? Take \(g=10\,\mathrm{m\,s}^{-2}\).
(a) For \(B\): \(5g-T=5a\). For \(A\): \(T=3a\). Add: \(50=8a \Rightarrow a=6.25\,\mathrm{m\,s}^{-2}\). \(T=3\cdot6.25=18.75\,\mathrm{N}\).
(b) After \(2\,\mathrm{s}\): speed of \(A\) \(=6.25\cdot2=12.5\,\mathrm{m\,s}^{-1}\). When string breaks, \(A\) has no force (smooth table) so continues at constant speed — it does not come to rest. (If the table were rough this would change; on smooth table \(A\) travels indefinitely at \(12.5\,\mathrm{m\,s}^{-1}\).)
Note: If the question intends a rough table, additional data would be needed. On a smooth table the answer is that \(A\) does not stop.
A particle of mass \(4\,\mathrm{kg}\) is projected up a rough plane inclined at \(30^\circ\) to the horizontal with initial speed \(8\,\mathrm{m\,s}^{-1}\). The coefficient of friction is \(0.2\). Find the distance the particle travels up the plane before coming to instantaneous rest, and the time taken. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Forces down the plane: \(mg\sin30^\circ+\mu mg\cos30^\circ=4\cdot10\cdot0.5 + 0.2\cdot4\cdot10\cdot\dfrac{\sqrt{3}}{2}=20+4\sqrt{3}\approx26.93\,\mathrm{N}\).
Acceleration down plane \(a=\dfrac{26.93}{4}\approx6.73\,\mathrm{m\,s}^{-2}\).
\(v=0\): \(0=8^2+2(-6.73)s \Rightarrow 45.36s\approx64 \Rightarrow s\approx1.41\,\mathrm{m}\).
Time: \(0=8-6.73t \Rightarrow t\approx1.19\,\mathrm{s}\).