A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 45.6 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding?
To find the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding, we can use the laws of motion and equations of motion.
First, let's analyze the forces acting on the crate when the truck is decelerating. There are two main forces:
1. The force of friction (μN) between the crate and the truck, where μ is the coefficient of friction and N is the normal force (equal to the weight of the crate).
2. The force of inertia acting on the crate due to its initial velocity.
Since the crate is at rest relative to the truck, the net force on the crate must be zero, meaning that the friction force must be equal to or greater than the force of inertia.
The force of inertia, F_inertia, can be calculated using Newton's second law of motion:
F_inertia = m⋅a
where m is the mass of the crate and a is the acceleration of the truck (which is equal to the deceleration since the crate is at rest relative to the truck).
Now, let's calculate the acceleration using the information given:
Initial velocity, u = 17.9 m/s
Final velocity, v = 0 m/s
Distance, s = 45.6 m
Using one of the equations of motion,
v^2 = u^2 + 2as
we can solve for the acceleration (a):
0^2 = (17.9 m/s)^2 + 2a(45.6 m)
Simplifying the equation:
320.41 m^2/s^2 = 91.2 a
a = 320.41 m^2/s^2 / 91.2
a ≈ 3.51 m/s^2
Now, we can calculate the force of inertia:
F_inertia = m⋅a
where m is the mass of the crate. Unfortunately, we don't have the mass of the crate given in the problem. Therefore, we cannot calculate the exact force of inertia at this point.
However, we can proceed with finding the minimum coefficient of friction. To do so, we can consider the force of friction (μN) and set it equal to the force of inertia (F_inertia):
μN = F_inertia
Since the crate is at rest relative to the truck, the normal force (N) is equal to the weight (m⋅g) of the crate, where g is the acceleration due to gravity.
So, we have:
μ⋅m⋅g = m⋅a
The mass (m) cancels out, giving:
μ⋅g = a
Now, we can isolate the coefficient of friction (μ):
μ = a / g
Substituting the known values:
μ = 3.51 m/s^2 / 9.8 m/s^2
μ ≈ 0.36 (rounded to two decimal places)
Therefore, the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding is approximately 0.36.