Equilibrium & Friction Forces: What's the Answer?

AI Thread Summary
An object in equilibrium has zero acceleration, meaning it can move at constant speed but has no net force acting on it. The discussion clarifies that while static friction is determined by the coefficient of static friction and the normal force, the contact area does not affect frictional force calculations. It is emphasized that static friction can vary to prevent motion, while the static friction force is not always equal to the coefficient times the normal force. The participants confirm the correctness of the answers provided and clarify misconceptions regarding friction forces. Understanding these principles is crucial for solving related physics problems effectively.
allstar1
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I need help with these 2 force questions. My answers have ***** next to them.

1) An object in equilibrium
Choices:
a) does not move.
b) has no forces acting on it.
c) has 0 accelleration. ***************
d) two of the above
e) all of the above

An object in equilibrium has to have a=0. I don't think it's choice A, because an object can move with constant speed and still be in equlibrium. I don't think it's choice b, because the net force equalls 0, but that doesn't mean that no forces are acting on it.



2) The static friction force
Choices:
a) equals the coefficient of static friction times the normal force.
b) is always less than the kinetic friction force.
c) will vary in order to keep the object from moving. ***************
d) depends on the area of contact between the surfaces.
e) is in the direction of motion.


choice B and E are the opposite of the truth. For A i think it's supposed to be less and not equal and I'm not really sure about choice D.
 
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You got them both right. Good job! :smile:

As for that second question, it is the coefficient of static friction (which quantifies the relative roughness) between two surfaces that determines the maximum static frictional force. The contact area has nothing to do with it. You can tell that that is wrong because you never use the contact area to calculate frictional forces.
 
Thanks for the reply.
 
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