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

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

Step-by-step 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 = (a)/(g)`

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

`\mu = (6.9)/(9.8)`

`\mu= 0.7667`

Answer 2

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