# Static/Kinetic Friction, Probelm solving question Purely algebraic

1. Nov 15, 2011

### Prodigium

1. The problem statement, all variables and given/known data
A rocket Propelled Trolly begins at rest time t = 0s, and then accelerates along a straight track such that the speed at time t is
$v(t)=bt^{2}$
where b is a constant, during the period 0<t<t2. at time $t_{2}$, the rocket fuel is exhausted and the trolley continues with constant speed
$v_{f}=bt^{2}_{2}$
A coin is initially at rest on the floor of the trolley. At time $t_{1}$, where $0<t_{1}<t_{2}$, it starts to slide backwards. It stops sliding at $t_{3}$, where $t_{3}>t_{2}$.
Use this information to obtain expressions for the coefficients of static and kinetic friction between the coin and the floor of the trolley.

2. The attempt at a solution
$a=\frac{d}{dt}(v(t))$

$a=2bt$

$F_{s}=\mu_{2}N$

$F_{s}=Ma$ (M=mass of trolley)

$N=mg$ (m=mass of coin)

$\mu_{s}=\frac{F_{s}}{N}$

$\mu_{s}=\frac{2btM}{mg}$

Thats as far as I got and I'm not even sure if what I've done is correct.

2. Nov 15, 2011

### cepheid

Staff Emeritus
Welcome to PF, Prodigium

Initially, when the cart begins to accelerate, Newton's 1st law would suggest that, in the absence of any external forces, the coin would remain stationary, which would mean it would begin to slide backwards relative to the floor of the cart. However, there ARE forces acting on the coin, namely static friction (which is forward facing, because it prevents the coin from sliding backwards relative to the cart floor). As a result, the coin accelerates along with the cart, and hence it does not slide relative to the cart floor. Therefore, I would say that in order for the coin to have the same acceleration a(t) as the cart does, the net force on the coin must be equal to:

F = ma(t)

note the lowercase m, not uppercase like you had. This is just Newton's 2nd Law. Unfortunately, this lack of sliding can only occur up to a limit, because the maximum amount of static frictional force that is available is equal to (the coefficient of static friction) * (the normal force) i.e.:

Fs ≤ μsN

Now, we know that sliding starts at time t = t1, and hence a = 2bt1. This is the instant at which the force required to accelerate the coin is equal to the maximum available static frictional force. Beyond this time, static friction will no longer be sufficient to prevent the coin from sliding. So, we equate the net force at this instant to the max value of static friction:

ma(t1) = μsN

2mbt1 = μs(mg)

μs = (2bt1)/g

Last edited: Nov 15, 2011
3. Nov 16, 2011

### Prodigium

Ah so because the force is acting on the coin not the trolley it's "m" instead of "M" and therefore cancels. Thanks now to attempt the kinetic coef.

4. Nov 16, 2011

### Prodigium

Thanks for the welcome, so far from what I've seen it's a brilliant site.

5. Nov 17, 2011

### Prodigium

could you give me a hint for the kinetic because im completley stuck and have been for a while. thanks