(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 wet concrete where the coefficient of kinetic friction is 0.5 and the coefficient of static friction is 0.7. (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) If the truck has four-wheel drive, and the cabinet is wooden, what is it's maximum acceleration (in m/s2)?

a)     a = 27.44 m / s²,  b) a = 5.39 m / s², c)  a = 156.8 m / s², cabinet maximum acceleration does not change

Explanation:

a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.

Let's apply Newton's second law

we set a regency hiss where the x axis is in the direction of movement of the truck

Y axis y

        N- W = 0

        N = W = m g

X axis

       2fr = m a

the expression for the friction force is

      fr = μ N

      fr = μ m g

we substitute

      2 μ m g = m /2   a

     a = 4 μ g

      a = 4 0.7 9.8

      a = 27.44 m / s²

b) let's look for the maximum acceleration that can be applied to the cabinet

       fr = m a

       μ N = ma

       μ m g = m a

       a = μ g

       a = 0.55  9.8

       a = 5.39 m / s²

as the acceleration of the platform is greater than this acceleration the cabinet must slip

c) the friction force is in the four wheels as well

With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all

applying Newton's second law

         4 fr = (m/4) a

         16 mg = (m) a

         a = 16 g

         a = 16 9.8

         a = 156.8 m / s²

cabinet maximum acceleration does not change

answer; /// i believe that the correct answer is (c)