# Elusive force opposing friction between passenger and seat in a decelerating car

A car traveling at 20m/s stops a distance of 50m. Assume deceleration is constant. The coefficient of static friction is 0.5 and coefficient of dynamic friction in 0.03. Will a 70kg passenger slid off the seat if not wearing a seat belt?

What I've done so far is find the acceleration (using kinematic eqn: $v^2 = u^2 + 2as$ where $v$ = final velocity and $u$ = initial velocity). I calculated deceleration to be $-4ms^{-2}$.

I also calculated force of static friction to be $350N$ ($F_s = (70kg)(10m/s^2)(0.5) = 350N$). Where I'm stuck is that when I made the free body diagram I can't determine what is the force opposing the friction between the passenger and the chair (since the car and the passenger are moving at a constant velocity before deceleration).

My guess is that you calculate the opposing force by multiplying the mass of the person by the deceleration ($F = ma = (70kg)(4ms^{-2}) = 280N$). However, I am not sure as to why this would work.

My other guess is that we calculate the kinetic energy of the passenger ($E_K = 1/2(70kg)(20ms^{-1})^2 = 14000J$). Since the "insert force here" is converting this kinetic energy to some other form of energy (thermal? sound?) as the car decelerates, therefore there is work being done ($14000J$ of work). I then equate $14000J$ to $Fs \cos\theta$ and solve for force and this gives me the same answer as my other method. ($14000J = Fs \cos\theta = F(50m)(\cos\theta$) Solving for $F$: $F = 280N$). However, this method also has holes in it.

• Addition to method two: is this force like a. inertial fictitious force akin to centrifugal force? – Aniekan Umoren Mar 19 '17 at 4:11

When the passenger is moving at constant velocity there is no frictional force on the passenger.

When the brakes are applied the seat (and the car) slows down from $20\, \rm m\,s^{-1}$ with an acceleration (I advise you not to use the tern deceleration) of $-4\,\rm m\,s^{-2}$.

The passenger also was moving at $20\, \rm m\,s^{-1}$ when the car started to slow down.

To slow the passenger down with an of acceleration $-4\,\rm m\,s^{-2}$ there must be a force of $280 \,\rm N$ acting on the passenger in a direction opposite to that of the velocity of the passenger.

So you now have the three forces that might be acting on the passenger.

• weight of the passenger - downwards
• force of seat on passenger upwards - normal reaction
• force of friction in opposite direction to the velocity of the passenger.

Remembering that for static friction the maximum force which could be applied is $\mu_{\rm s} N$ where $\mu_{\rm s}$ is the coefficient of static friction and $N$ is the normal reaction you have to decide if the static friction is large enough to accelerate the passenger at $-4\,\rm m\,s^{-2}$.
Note the word maximum.
In other words the static friction force can vary between $0$ and $\mu_{\rm s}$ depending on the situation.
As I stated before the static frictional force is zero when the car and the passenger are travelling at constant velocity.
So there is no "elusive" force opposing friction in this situation as the static frictional force is zero.

If it is not then there will be relative movement between the passenger and the seat and the kinetic frictional force acting in a direction opposite to the passenger's velocity will slow the passenger down at a smaller rate than the car.

• ooohhhh we have to see whether force of static friction can be greater than or equal to 280N since this is the force required to accelerate the passenger by -4m/s. Right? – Aniekan Umoren Mar 19 '17 at 16:44