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A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.2 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?

Respuesta :

Answer:

   μ < 0.73

Explanation:

Given that

Acceleration ,a= 7.2 m/s²

Lets take mass of the car = m

The coefficient of friction = μ

We know that friction force fr

fr= μ m g

From Newtons second law

F= m a

By balancing the forces

m a = μ m g

a =  μ g

[tex]\mu=\dfrac{a}{g}[/tex]

[tex]\mu=\dfrac{7.2}{9.81}[/tex]

μ= 0.73

So if coefficient of friction < 0.73 then the car will slip.

μ < 0.73

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