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Two objects on an incline beside each other, same coefficient of friction, #1 box weight=1 kg, the other #2 box=1000 kg. As the incline rises and the maximum static friction force is exceeded the boxes will slide down the incline. Will they both slide at the same angle of inclination independent of their masses. Does having the same coefficient of friction, which is independent of mass, have the same maximum static frictional force? Wouldn't the mass reflect in the max static frictional force and therefore each box move at a different angle of inclination and yet still have the same coefficient of friction? This is not about acceleration just the initial move of each box from its stationary position.

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The heavier box has 1000 times larger maximum static friction force than the lighter box, as you note (this holds for every angle). However, the heavier box also pushes against this friction 1000 times harder (for any angle). Therefore they will both start moving at the same angle.

(In reality such a large difference in mass world likely break the assumption of a single common friction coefficient. The heavier box may for example well make an indentation for itself that makes it harder for it to start moving. Or it could go the other way.)

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    $\begingroup$ Thank you Kristoffer! This has helped my understanding clearly. $\endgroup$
    – C Lake
    May 15 at 12:49
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Will they both slide at the same angle of inclination independent of their masses.

Yes. Impending motion of both will occur at the same angle and that angle will satisfy the following equation where $\mu_s$ is the coefficient of static friction.

$$\tan\theta=\mu_s$$

PROOF: (See free body diagram, below)

The gravitational force acting down the incline is

$$mg\sin\theta$$

The maximum possible static friction force acting up the plane is

$$F_{f}=\mu_{s}N=\mu_{s}mg\cos\theta$$

Impending motion occurs when the two are equal, or

$$mg\sin\theta=\mu_{s}mg\cos\theta$$

$$\mu_{s}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

Hope this helps.

enter image description here

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    $\begingroup$ Thank you Bob! The formula and diagram help put it all together. $\endgroup$
    – C Lake
    May 15 at 12:50

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