5. A 7.6 kg object is at rest on an inclined plane. If the inclined plane makes an angle with the horizontal

of 33', what is the coefficient of static friction between the object and the plane? Assume the box is just

about to slide and this is the marimum static friction.

Correct answer: Cf = tan 33° = 0.649 (I guess you meant degrees and not minutes)

Explanation:

The horizontal component of the object's weight equals to the force of friction and the object is just about to slide.

The horizontal component of the object's weight is Qh = mg sin 33°

The force of friction is Ffr = Cf · Qv = Cf · mg cos 33°, where Cf is the coefficient of static friction and Qv vertical component of the object's weight

Qh - Ffr = 0 = > Qh = Ffr = > mg sin 33° = Cf mg cos 33°

we divide both sides of the equation by mg and get:

Cf · cos 33° = sin 33° = > Cf = sin 33° / cos 33° = tan 33° = 0.649

Cf = 0.649 the weight of the object is unnecessary data

God is with you!

Answer:0.65

Explanation:

Since the box is about to slide down

coefficient of friction = tanβ

Coefficient of friction = tan33

Coefficient of friction = 0.65