A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide?

I've tried this and I still can't get it. Can someone please show me the steps?

Max friction force = M g u

The equals M a when it is on the verge of slipping.
Therefore a = u g is the maximum deceleration rate. u is the static coefficient of friction, 0.27
a = 2.646 m/s^2
To get the stopping distance X, use
V = sqrt (2 a X)= sqrt (2 u g X)
Convert the V = 59.7 km/hr to m/s units before using that equation.

To determine the minimum stopping distance for the train without causing the crates to slide, we need to consider the forces acting on the crates. The maximum force of static friction (Fs) between the crates and the floor is given by the equation:

Fs = μs * N

where μs is the coefficient of static friction and N is the normal force exerted on the crates. In this case, the normal force is equal to the weight of the crates since there is no vertical acceleration. Therefore, we can rewrite the equation as:

Fs = μs * m * g

where m is the mass of the crates, and g is the acceleration due to gravity.

The force required to stop the train is equal to the product of the mass of the train (M) and the acceleration (a):

F = M * a

Since the force of static friction provides the required force to stop the train, we can equate these two forces:

Fs = F

μs * m * g = M * a

Next, we need to convert the velocity of the train from km/hr to m/s:

59.7 km/hr = (59.7 * 1000) / 3600 = 16.583 m/s

Now, we can calculate the deceleration (a) using the formula:

a = (v^2) / (2 * d)

where v is the initial velocity (16.583 m/s), and d is the stopping distance.

Rearranging the formula, we get:

d = (v^2) / (2 * a)

Substituting the given values, we have:

d = (16.583^2) / (2 * a)

Now, we can substitute the expression for acceleration (a) from the equation μs * m * g = M * a:

d = (16.583^2) / (2 * (μs * m * g / M))

Finally, we substitute the given values for the coefficient of static friction (μs = 0.27), the mass of the crates (m), the acceleration due to gravity (g = 9.8 m/s^2), and the mass of the train (M):

d = (16.583^2) / (2 * (0.27 * m * 9.8 / M))

Simplifying this equation will give you the minimum stopping distance (d) required to stop the train without causing the crates to slide.