A package of dishes (mass 50.0 kg) sits on the flatbed of a pickup truck with an open tailgate. The coefficient of static friction between the package and the truck's flatbed is 0.330, and the coefficient of kinetic friction is 0.230.

(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the package does not slide relative to the truck bed?




(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the package sliding along its bed. What is the acceleration of the package relative to the ground?


(c) The driver cleans up the fragments of dishes and starts over again with an identical package at rest in the truck. The truck accelerates up a hill inclined at 9.0° with the horizontal. Now what is the maximum acceleration the truck can have such that the package does not slide relative to the flatbed?


(d) When the truck exceeds this acceleration, what is the acceleration of the package relative to the ground?


(e) For the truck parked at rest on a hill, what is the maximum slope the hill can have such that the package does not slide?
°

(f) Is any piece of data unnecessary for the solution in all the parts of this problem?
The coefficient of static friction
The mass of the package
The coefficient of kinetic friction

To answer these questions, we need to utilize the frictional forces acting on the package. Let's go step by step.

(a) The maximum acceleration the truck can have so that the package does not slide relative to the truck bed can be found by equating the maximum static frictional force to the maximum force of friction before sliding occurs. The formula for static friction is given by:

F_s = µ_s * N

where F_s is the static frictional force, µ_s is the coefficient of static friction, and N is the normal force exerted on the package by the truck bed.

To find N, we need to consider the weight of the package. The weight is given by:

W = m * g

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

Therefore, N = W = m * g.

Now we can substitute N into our earlier equation for static frictional force, F_s = µ_s * N, and equate it to the limit given by the equation F_s = m * a, where a is the acceleration of the truck. This gives us:

µ_s * m * g = m * a

Solving for a, we find:

a = µ_s * g

Plugging in the given values, we get:

a = 0.330 * 9.8 m/s²
a ≈ 3.234 m/s²

Therefore, the maximum acceleration the truck can have without the package sliding relative to the truck bed is approximately 3.234 m/s².

(b) Now, the truck barely exceeds this acceleration, so the package starts to slide relative to the truck bed. The force acting to oppose the motion is kinetic friction. The formula for kinetic frictional force is given by:

F_k = µ_k * N

where F_k is the kinetic frictional force, µ_k is the coefficient of kinetic friction, and N is the normal force exerted on the package by the truck bed.

Similar to part (a), N = m * g.

Therefore, the force equation becomes:

µ_k * m * g = m * a'

where a' is the acceleration of the package relative to the ground.

Solving for a', we get:

a' = µ_k * g

Plugging in the given values, we have:

a' = 0.230 * 9.8 m/s²
a' ≈ 2.254 m/s²

Therefore, the acceleration of the package relative to the ground is approximately 2.254 m/s².

(c) Now, the truck accelerates up a hill inclined at 9.0° with the horizontal. We need to consider the components of forces acting on the package along the incline. The normal force exerted on the package by the truck is given by:

N = m * g * cos(θ)

where θ is the angle of the incline.

The static frictional force acting along the incline is given by:

F_s = µ_s * N

Since the truck just accelerates without the package sliding, we can equate this static frictional force to the gravitational force component along the incline, which is:

F_g = m * g * sin(θ)

Therefore, we have the equation:

µ_s * m * g * cos(θ) = m * g * sin(θ)

Simplifying and solving for θ, we get:

θ = arctan(µ_s)

Plugging in the given value of µ_s, we have:

θ ≈ arctan(0.330)
θ ≈ 17.98°

Therefore, the maximum acceleration the truck can have without the package sliding relative to the flatbed is when the hill is inclined at an angle of approximately 17.98°.

(d) When the truck exceeds this acceleration, the package starts to slide relative to the flatbed. The force opposing the motion is kinetic friction, and the acceleration of the package relative to the ground is the same as the acceleration of the truck.

So, the acceleration of the package relative to the ground is the same as the acceleration of the truck in part (c), which is approximately 3.234 m/s².

(e) For the truck parked at rest on a hill, the maximum slope the hill can have such that the package does not slide can be calculated using the formula found in part (c):

θ = arctan(µ_s)

Plugging in the given value of µ_s, we get:

θ ≈ arctan(0.330)
θ ≈ 17.98°

Therefore, the maximum slope the hill can have without the package sliding is approximately 17.98°.

(f) The coefficient of static friction, the mass of the package, and the coefficient of kinetic friction are all necessary for solving all the parts of this problem. Hence, none of the given data is unnecessary for the solution in all parts.