A. If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve on dry concrete where the coefficient of kinetic friction is 0.7 and the coefficient of static friction is 1m/s^2. b. Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate where the coefficient of kinetic friction is 0.3 and the coefficient of static friction is 0.55?
c. Answer both of these questions for the case that the truck has four-wheel drive, and the cabinet is wooden.

A) a_max = 4.9 m/s²

B) the metal cabinet will not slip.

C) a_max = 9.8 m/s². The cabinet will slip.

Explanation:

A) We are given;

Coefficient of static friction; μ_s = 1

Coefficient of kinetic friction; μ_k = 0.7

Formula for maximum static friction;

F_s = μ_s•N

We are dealing with half of the weight of a flatbed truck. Thus;

N = mg/2

Thus;

F_s(max) = ½μ_s•mg

Now, the maximum acceleration it can achieve on dry concrete will be when the Maximum static friction is reached.

Because after maximum static friction, the cabinet will slip.

Thus; F_s(max) = ma

Therefore,

ma = ½μ_s•mg

m will cancel out to give;

a = ½μ_s•g

a = ½ × 1 × 9.8

a_max = 4.9 m/s²

B) we are told that the coefficient of static friction is now 0.55

Thus, the friction between the metal and the wood is;

F = μ_s•g = 0.55 × 9.8

F = 5.39 N

The acceleration gotten in the first part is less than μ_s•g = 5.39 N, then the metal cabinet will not slip.

C) Now, if we are considering a 4 wheel drive, then we will not divide the mass by 2 and so; N = mg

Like we did in A above;

ma = μ_s•mg

a = μ_s•g

a = 1 × 9.8

a = 9.8 m/s²

Now, this value of a_max is greater than μ_s•g in answer A above.

Thus, a_max > μ_s•g and thus, the cabinet will slip.

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