The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.34 with the floor. If the train is initially moving at a speed of 58 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

To determine the shortest stopping distance of the train without causing the crates to slide over the floor, we need to consider the forces acting on the crates.

1. Determine the deceleration of the train:
First, convert the initial speed of the train from km/h to m/s.
58 km/h = 58,000 m/3600 s = 16.11 m/s

Since the train is brought to a stop, the final velocity is 0 m/s.

Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can rearrange it to solve for acceleration (a):

0^2 = (16.11 m/s)^2 + 2a · s
0 = 259.92 m^2/s^2 + 2a · s
-2a · s = 259.92 m^2/s^2
a · s = -259.92 m^2/s^2
a = -259.92/s m/s^2

Since we are assuming constant acceleration, the negative sign indicates deceleration.

2. Determine the maximum friction force:
The maximum friction force (F_friction) that can be exerted on an object is given by the coefficient of static friction (μ_s) multiplied by the normal force (F_normal) acting on the object. In this case, the normal force is equal to the weight of the crates (F_gravity).

Assuming the mass of the crates is m, the weight of the crates can be calculated as F_gravity = m · g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

3. Calculate the stopping distance:
The friction force acting on the crates opposes the motion and contributes to the deceleration. When the frictional force reaches its maximum value, the crates begin to slide.

Since the friction force (F_friction) is equal to the mass of the crates (m) multiplied by the acceleration (a), it must also be equal to the maximum friction force (F_max_friction).

F_friction = F_max_friction
m · a = μ_s · m · g

Canceling out the mass (m) on both sides of the equation, we get:
a = μ_s · g

Now we can substitute this value for a into the equation relating acceleration, distance, and initial speed:
a = -259.92/s m/s^2
-259.92/s = 0.34 · 9.8 m/s^2
-259.92/s = 3.332 m/s^2

Solving for s:
s = -259.92 m/s^2 / 3.332 m/s^2
s ≈ 78.01 m

Therefore, the train can be stopped in approximately 78.01 meters without causing the crates to slide over the floor.