Rich example

## Solution

A particle of mass $\quantity{10}{kg}$ starts at rest on a rough inclined plane. A force of magnitude $\quantity{P}{N}$ is applied to the particle and acts up the slope, parallel to the plane.

When $P = 20$ the particle is on the point of sliding down the plane.

When $P = 40$ the particle is on the point of sliding up the plane.

Before we can tackle the problems below we need to think about the situation we have been given and what information we know. For both values of $P$ the particle is at rest, so the forces acting on it are in equilibrium.

Why is the frictional force going in different directions on each diagram?

What do we know about frictional force when a particle is on the point of moving?

Resolving the forces parallel to the plane for each diagram gives us the following equations:

The model we use for frictional force is $F_r ≤ \mu R$, so we have to be careful when writing down equations that we don’t assume that any $F_r$’s appearing will represent the same value. In this case, both particles are on the point of moving, so $F_r= \mu R$ for both situations.

As $F_r$ is the maximum frictional force in both cases we can write

\begin{align*} 20+F_{max} &= 10g\sin\theta \\ 40 - F_{max} &= 10g\sin\theta \end{align*}

where $F_{max} = \mu10g\cos\theta$.

We can solve the equations to find that $F_{max} = \quantity{10}{N}$ and $10g\sin\theta = \quantity{30}{N}$.

What does the force $10g\sin\theta = \quantity{30}{N}$ represent?

Do we need to calculate $\theta$?

For each of the following values of $P$ describe how the particle is moving and calculate the frictional force acting on it:

• $P = 25$

As we are about to compare values, we need to be careful about the direction of any forces we calculate. We will take frictional force to be positive as it goes up the slope.

When $P = 25$ the particle must remain at rest as the value of $P$ is between $20$ and $40$. Therefore the forces are in equilibrium and $25+F_r = 10g\sin\theta.$ We know $10g\sin\theta = 30$, so we find that $F_r = 5$ and is acting up the slope.

The frictional force could also be found by comparison. Compared to when $P=20$, the force of $P$ has increased by $5$, so the frictional force must have decreased by $5$ in order to keep the equilibrium.

• $P = 30$

When $P = 30$ the particle must remain at rest as the value of $P$ is between $20$ and $40$. Therefore the forces are still in equilibrium, so

$30+F_r = 10g\sin\theta,$

giving us $F_r = 0$.

What does a frictional force of $0$ mean? Is there another way that we could have reasoned that $F_r = 0$ when $P=30$?

• $P = 50$

When $P = 50$ the particle must be accelerating up the plane, since it is on the point of sliding up when $P=40$.

How do we know it’s accelerating? Do we know how fast?

As the particle is no longer in equilibrium we can apply Newton’s second law. \begin{align*} F&=ma \\ P + F_r - 10g\sin\theta &= ma \\ \end{align*}

The model we use for friction when a particle is moving is $F_r = \mu R$, which is the maximum frictional force of $\quantity{10}{N}$ that we found earlier. However, this force is acting down the slope so we will think of this as a negative value.

\begin{align*} 50 - 10 - 30 &=10a \\ a &= \quantity{1}{m\,s^{-2}} \end{align*}

What would a graph of the frictional force against $P$ look like?

In order to sketch the graph, or check that your sketch seems reasonable, it might be helpful to think about the following questions:

• What happens to friction between $P=20$ and $P=40$?
• Will the graph be linear or curved? Why?
• Does the graph fit with our model of friction, $F_r ≤ \mu R$?

Would knowing what the graph looked like have helped you to find the frictional force for each value of $P$ in a different way?