Question

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?

Answer 1

Answer:

The value is  $\mu = 0.76$

Explanation:

From the question we are told that

    The  acceleration is $a = 7.4 \ m /s^2$

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

          $F = ma$

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

        $F_f = F$

Here  $F_f$ is the frictional force which is mathematically represented as

         $F_f = \mu * m * g$

substituting into above equation

         $\mu * m * g = ma$

=>        $\mu = \frac{a}{g}$

substituting values

           $\mu = \frac{ 7.4 }{ 9.8}$

           $\mu = 0.76$