A crate of mass 52.8 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the trucks flatbed is 0.351, and the coefficient of kinetic friction is 0.305.
(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the trucks flatbed?
(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?
Huge shoes^3
To answer these questions, we need to apply the concept of friction and Newton's second law of motion. Let's break it down step by step.
(a) The maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed is determined by the force of static friction. The formula for static friction is given by:
Fs ≤ µs * N
where:
- Fs is the force of static friction
- µs is the coefficient of static friction
- N is the normal force
In this case, the normal force is equal to the weight of the crate, which is given by:
N = m * g
where:
- m is the mass of the crate
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values, we have:
N = 52.8 kg * 9.8 m/s^2
N ≈ 517.44 N
Now, we can calculate the maximum force of static friction:
Fs ≤ 0.351 * 517.44 N
Fs ≤ 181.17 N
Using Newton's second law (F = m * a), we can equate the static friction force with the force required to accelerate the crate:
Fs = m * a
181.17 N = 52.8 kg * a
Solving for acceleration, we get:
a = 181.17 N / 52.8 kg
a ≈ 3.43 m/s^2
Therefore, the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed is approximately 3.43 m/s^2.
(b) Once the truck exceeds this maximum acceleration and the crate starts to slide, the force of kinetic friction comes into play. The formula for kinetic friction is similar to static friction but uses the coefficient of kinetic friction (µk) instead:
Fk = µk * N
Using the same procedure as in part (a), we find the force of kinetic friction:
Fk = 0.305 * 52.8 kg * 9.8 m/s^2
Fk ≈ 152.17 N
Since the crate is sliding, the net force acting on it is given by:
ΣF = m * a
Taking into account the force of kinetic friction:
152.17 N - m * g = m * a
Solving for acceleration, we get:
a = (152.17 N - 52.8 kg * 9.8 m/s^2) / 52.8 kg
a ≈ 1.23 m/s^2
Therefore, the acceleration of the crate relative to the ground, once the truck moves with constant acceleration and the crate slides along its bed, is approximately 1.23 m/s^2.