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?
Ok so I understand that mu*mass*g = mass*v^2/2d, and I need to solve for d. But how do I get the mass?
To calculate the mass of the crates, you can use the equation:
F_friction = µ * N,
where F_friction is the force of friction, µ is the coefficient of static friction, and N is the normal force. The normal force is equal to the weight of the crates, which is given by:
N = m * g,
where m is the mass of the crates and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the coefficient of static friction µ is 0.27, you can substitute these values into the equation:
F_friction = µ * N = µ * m * g.
The force of friction can also be expressed as the deceleration of the crates multiplied by their mass:
F_friction = m * a,
where a is the deceleration. Substituting this value into the equation:
m * a = µ * m * g.
Now, you can cancel out the masses on both sides of the equation:
a = µ * g.
Given that the deceleration a is equal to v^2 / (2 * d),
v^2 / (2 * d) = µ * g.
Now, rearrange the equation to solve for d:
d = v^2 / (2 * µ * g).
Substituting the given values, with v equal to 59.7 km/hr (or approximately 16.58 m/s) and µ equal to 0.27, and g equal to 9.8 m/s^2, you can calculate the distance d that the train can be stopped without causing the crates to slide.