Question

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.9 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 cabinet will slip along the bed surface for the values of coefficient of static friction lesser than 0.7

Explanation: Please see the attachments below

Answer 2

Answer:

The coefficient of static friction is  $\mu= 0.7667$

Explanation:

 From the question we are told that

          The acceleration is $a = 6.9 m/s^2$

The mathematical expression that show the relationship between frictional force and the sliding  force of the cabinet at the point just before sliding begins is

                                         $F_f = F_s$

Now  $F_f = mg\mu$

   Where m is the mass of the cabinet

                g is the acceleration due to gravity

               $\mu$ is the coefficient of static friction

And $F_s = ma$  

  Where a is the acceleration due to gravity

               Now substituting this into the equation

                        $mg \mu = ma$

                        $g\mu =a$

                       $\mu = \frac{a}{g}$

Substituting values where $g =9.8m/s^2$

                  $\mu = \frac{6.9}{9.8}$

                    $\mu= 0.7667$