W4.2 Riding the bus
You are travelling home on the bus, having just bought a bottle of something rather nice for your friend’s birthday, which is sitting on the seat next to you, in a plastic bag. It’s a typical rainy day in Edinburgh and the bus is late so the driver is going quite fast to make up time. At a junction, the driver has to slam on the brakes, locking up the wheels and bring the bus to a halt in 42m from an initial speed of 65 kmh.
You are sitting at the back of the bus and are taken by surprise by the sudden stop. You don’t have time to make a grab for the present. Will it slide off the seat?
Hint
Reveal

Hint
Here’s some data. You may or may not need to use all of these:
The mass of the bus plus passengers is 10,000 kg
Coefficient of static friction for: rubber on wet roads (0.5) and plastic bags and vinyl bus seats (0.35)
Coefficients of kinetic friction for: rubber on wet roads (0.4) and plastic bags on vinyl bus seats (0.25)
You know the procedure by now... think FDPEE....
Solution
Reveal

Solution
This in some ways is a backward problem; you will be used to doing problems where the ’natural state’ of the object is at rest, and something acts on it to cause it to accelerate and thus move. This is the other way round - the natural state of the present next to you is to move at 65 kmh (a constant speed). What happens here is that a force acts on it causing it to come to a stop.
Let’s start from the free body diagram for the present while the bus is moving at constant speed. There’s no horizontal force on the present: it is not accelerating but moving at a constant speed. (The vertical forces, weight and normal contact force are balanced too)
Let’s imagine the bus is travelling to the right in your mental picture of the problem. When the bus brakes, it is subject to an acceleration to the left. The FBD for the box now must include the frictional force of the seat on the present, and this will act to the left (so as to oppose the naturtal motion of the box, (which is to carry on moving forwards!)
The crux of this problem is to find if the maximum acceleration of the present (to the left) that can be provided by the frictional force is more than the acceleration to the left that the bus undergoes. If this is the case, the present stays put. If not, it slides and you’re probably off to buy a replacement.
So our strategy is:
(1) Find the acceleration of the bus from the speed and distance information given (assuming it is constant)
(2) Find the maximum value of the frictional force on the present (which occurs at the point of slip)
(3) Thus find the maximum acceleration of the present
(4) Compare these two accelerations and arrive at a judgement.
So:
(1) We can use the constant acceleration equation to calculate ab=-3.9,s-2. (Note that you need to convert 65 kmh to 18 ms-1)
(2) The maximum frictional force (on the point of slip) is given by Ff,max=μsFN=μsmg. (We’re using the static coefficient of friction here as we’re just on the point of slipping). This force acts to the left.
(3) For the box, we can now invoke Newton’s Second Law
We can calculate this value as -3.4 ms-2
(4) The maximum acceleration of the box is not enough for it to decelerate as quickly as the bus
It will slide off.