2D Forces & Friction

When a force is applied, and it is not pointed in the direction of motion, it can be resolved to find the component that does point in the direction of motion. For example, imagine when a person is sweeping the floor, the direction of motion is horizontal but a diagonal force is being applied – we can split this force into two components: the one that works horizontally and the one that works vertically. This is known as resolving a force and we can use trigonometry to do it (remind yourself of how to use SOHCAHTOA).

We can also use the triangle law for vector addition that we learned about in AS Maths to help find resultant forces – see Example 1.

Inclined Planes

Similarly, if the direction of motion is on an inclined plane, such as a ball rolling down a hill, we can resolve the forces that work vertically and horizontally (such as the weight of the ball) into the direction of the plane of motion and the one perpendicular to it. See Example 2.


Up until now, all of the forces working on a moving object modelled as a particle have been on a smooth plane. This means that we have been ignoring the force of friction which opposes motion when the surface has some roughness to it. The amount of roughness the surface possess is quantified by the coefficient of friction, usually called \mu. The coefficient of friction is 0 (\mu=0) for a smooth surface but for a rough surface \mu\ne 0 and the rougher the surface the larger the value of \mu.

Calculating the force of friction is dependent on the other forces in play. Imagine you would like to push a bag of potatoes along a rough floor (or up a rough plane). If you apply a gentle push of say 1 Newton, friction will oppose the direction of motion and will also be equal to 1 Newton. As you apply more force to move the bag of potatoes, friction ({\bf Fr}) will also increase to that force until it reaches its maximum value at which point the bag will begin to move. The maximum value of friction is {\bf Fr}=\mu{\bf R}, where {\bf R} is the reaction force – the one that stops an object from sinking into the surface it lies on. So, if an object is moving (or about to move) then {\bf Fr}=\mu{\bf R}, otherwise it is calculated from the other forces acting on the object to obtain equilibrium – see Example 3. Note that when an object has acceleration ({\bf a}), the usual equation of motion ({\bf F}=m{\bf a}) applies where friction is now included in the resultant force ({\bf F}).

Object not moving – friction is less than limiting value (equal to limiting value when it is about to move).
Object moving (either with zero or constant acceleration) – friction is equal to limiting value.

Friction can be applied in examples using connected particles, such as pulleys, when one of the particles is moving along a rough surface as in Example 3. Other examples may use friction acting on rigid bodies, as opposed to objects modelled as particles. These examples may also require the use of moments (see moments) – see Example 4.


Three forces act on a particle as shown in the diagram. Given that the particle is in equilibrium, find the magnitude of the force F to 2 decimal places.

Since the particle is in equilibrium, the horizontal components of the forces balance, i.e. the components of the forces that act to the left equal the components of the forces that act to the right. It follows that F\cos(\theta)+50\cos(45^\circ)=75\cos(60^\circ) and so F\cos(\theta)\approx 2.1447 (don’t round to 2 decimal places until the end). Similarly for the vertical components, the parts that act upwards equal the parts that act downwards. So, F\sin(\theta)=50\sin(45^\circ)+75\sin(60^\circ)=\approx 100.3072. To find the magnitude of F we use the fact that F^2=\left(F\cos(\theta)\right)^2+\left(F\sin(\theta)\right)^2 since \cos^2(\theta)+\sin^2(\theta)=1 where (F\cos(\theta))^2\approx 4.5996 and (F\sin(\theta))^2\approx 10061.5433. Hence F=\sqrt{4.5996+10061.5433}\approx 100.33 Newtons to 2 decimal places.

An object of mass 5kg is pushed up a smooth slope by a force of magnitude 30N which is acting an angle of 50^\circ to the slope. Given that the slope is at an angle of 40^\circ to the horizontal, and modelling the object as a particle, find the acceleration of the object.

We begin by sketching a diagram showing the relevant forces:

In this example, we can resolve the forces into components parallel to the slope. Note that the slope is smooth so there is no frictional force. The component of the force pushing upwards parallel to the slope is 30\cos(50^\circ)\approx19.28N and the component of weight pulling downwards is 5g\sin(40^\circ)\approx 31.50N. It follows that the object will accelerate down the slope with a resultant force of 12.22N. This force is equal to mass times acceleration (F=ma) and so the acceleration is given by \frac{12.22}{5}=2.44m/s/s.

Two particles A and B are connected by a light inextensible string. The particles have masses 3kg and 6kg respectively. Particle A sits on a rough plane inclined at an angle of 30^\circ to the horizontal, the string is taut and passes over a smooth light pulley at the top of the inclined plane and particle B hangs at the end of it. The coefficient of friction between particle A and the plane is 0.25 and the system is released from rest.
a) Find the acceleration of the particles and the tension in the string.
b) State how you have used the assumption that the string is inextensible.
c) Find the magnitude of the force exerted by the string on the pulley.

Again we begin by sketching a diagram showing the forces:

a) Assuming that acceleration is up the slope as particle B is heavier, we can construct the equations of motion for each of the particles separately. Starting with particle B (the simpler one), F=ma (where F is the resultant force) becomes 6g-T=6a or 58.8-T=6a using g=9.8m/s/s. Both T and a are unknown so we need a second equation (the equation of motion for particle A) to solve simultaneously. Particle A has the same acceleration and tension and the equation of motion is T-Fr-3g\sin(30^\circ)=3a where we need an expression for friction. The particle is moving so friction is working at its maximum Fr=\mu R where \mu=0.25 and R is equal to the component of weight that works perpendicularly to the slope. It follows that Fr=0.25\times 3g\cos(30^\circ)\approx 6.365N to 3 decimal places. The equation of motion for particle A simplifies to T-21.065=3a. We solve the equations of motion simultaneously by first eliminating T to obtain a=4.2m/s/s to 1 decimal place. It follows that T=33.6N to 1 decimal place.
b) We used the fact that the string is inextensible when we set the acceleration of both particles to be the same.
c) Isolating and treating the pulley as a particle itself, there are only two forces acting on it, both equal to the tension found in part a), but pulling the pulley towards the plane.

Since these forces are equal, the resultant force acts through the middle of the angle between them. The component of each force acting in that direction is T\cos(30^\circ)\approx 29.1N. It follows that the magnitude of the force exerted on the pulley by the string is 58.2N.
Note that there is an equal and opposite reactive force of the pulley on the string but we are only interested in the force of the string on the pulley here.

A ladder of mass x kg and length 3 metres stands on rough horizontal ground at an angle of 60^\circ to the horizontal and rests against a smooth vertical wall. A man of mass 10x kg is standing a third of the way up the ladder.
a) Modelling the man as a particle and the ladder as a uniform rigid body in equilibrium, find the reaction force at the bottom of the ladder in terms of x and g.
b) Find the reaction force at the top of the ladder.
c) Find the frictional force at the bottom of the ladder and show that the coefficient of friction between the ladder and the ground must satisfy \mu\geq \frac{16\sqrt{3}}{99}.
d) State how you have used the assumption that the ladder is uniform in your calculations.

Once again we sketch a diagram to show the forces:

a) Since the ladder is in equilibrium, the downward forces match the upward forces. It follows that the reaction force at the bottom of the ladder is R1=10xg+xg=11xg.
b) There is no suggestion that friction is working at its maximum so we don’t use Fr=\mu R1. So, in order to find the reaction force at the top of the ladder, we take moments around the point at the bottom (see more on moments). The clockwise moments are given by 1.5\times 10xg\cos(60^\circ)+1\times xg\cos(60^\circ)=8xg. The anticlockwise moment is given by 3\times R2\sin(60^\circ)=\frac{3\sqrt{3}}{2}R2. Since the moments balance we find that R2=\frac{8xg}{\frac{3\sqrt{3}}{2}}=\frac{16\sqrt{3}xg}{9}.
c) Horizontal forces balance and so the frictional force is also equal to the reaction at the top of the ladder, that is Fr=\frac{16\sqrt{3}xg}{9}. Frication is working at this value which must be less than or equal to \mu R1=11\mu xg. It follows that \frac{16\sqrt{3}xg}{9}\leq 11\mu xg and since both x and g are positive, \mu \geq \frac{16\sqrt{3}}{99}.
d) We use the assumption that the ladder is uniform by setting the centre of mass of the ladder to be acting at the centre of the ladder.