Paper 2 Forces Answers

Q1 Force resolution on an incline

1. Weight $mg$ acts vertically downward; normal contact force acts perpendicular to the plane. Since the plane is smooth, no friction acts. [2]
2. Parallel component $=mg\sin 18^{\circ }=5.0(9.81)\sin 18^{\circ }=15.2\mathrm{N}$ down the plane. [2]
3. Normal force $=mg\cos 18^{\circ }=5.0(9.81)\cos 18^{\circ }=46.6\mathrm{N}$. [2]
4. Weight is the gravitational force exerted by Earth on the box. The normal force is exerted by the plane on the box. They act on the same body, so they are not a third-law pair; the partner of the weight is the gravitational force exerted by the box on Earth. [2]

Q2 Hooke’s law

1. Convert $6.0\mathrm{cm}=0.060\mathrm{m}$. $k=F/x=3.6/0.060=60\mathrm{N}{\mathrm{m}}^{-1}$. [2]
2. $x=F/k=4.8/60=0.080\mathrm{m}=8.0\mathrm{cm}$. [2]
3. The limit of proportionality is the point up to which force is directly proportional to extension, so $F/x$ is constant. [2]
4. Beyond the limit, the material no longer responds with constant stiffness; the ratio $F/x$ changes, so the force-extension graph is no longer a straight line. [2]

Q3 Moments and equilibrium

1. For rotational equilibrium, total clockwise moment equals total anticlockwise moment about any pivot. [1]
2. Anticlockwise moment from left load $=12.0(0.40)=4.80\mathrm{N}\mathrm{m}$. For equilibrium, $W(0.30)=4.80$, so $W=16.0\mathrm{N}$. [3]
3. The beam is uniform, so its weight acts through its centre. The pivot is also at the centre, so the perpendicular distance from the pivot to the weight’s line of action is zero. [2]
4. The resultant force must also be zero. [1]