- #1
Beamsbox
- 61
- 0
I just realized I don't even have to do this problem, but I've thought too much about it, and at times like this I tend to be a bit OCD.
If someone could just get me started on the right track of thought that would be great!
Q:
In figure:
(http://i51.photobucket.com/albums/f362/BeamsBox/PhysicsBlocks.jpg)
a horizontal force of 100N is to be applied to a 10kg slab that is initially stationary on a frictionless floor, to accelerate the slab. A 10kg block lies on top of the slab; the coefficient of friction (mue) between the block and the slab is not known, and the block might slip.
A) Considering the possibility, wha tis the possible range of values for the magnitude of the slab's acceleration aslab? (Hint: You don't need written calculations; just consider extreme values for (mue).)
My thoughts:
First, the force of friction between the slab and the block,
Fs(max) = (mue)FN, and
FN = mg.
So,
Fs(max) = (mue)(mg)
Now, noting this relation, it makes sense to me that Fs(max) cannot be greater than FN... but this could also be wrong, because it's not that it can't be greater than FN, it just can't be greater than FN times (mue)... which is unknown... that's where I lose myself.
I just don't know what I need to consider, that I'm obviously not...
;)
If someone could just get me started on the right track of thought that would be great!
Q:
In figure:
(http://i51.photobucket.com/albums/f362/BeamsBox/PhysicsBlocks.jpg)
a horizontal force of 100N is to be applied to a 10kg slab that is initially stationary on a frictionless floor, to accelerate the slab. A 10kg block lies on top of the slab; the coefficient of friction (mue) between the block and the slab is not known, and the block might slip.
A) Considering the possibility, wha tis the possible range of values for the magnitude of the slab's acceleration aslab? (Hint: You don't need written calculations; just consider extreme values for (mue).)
My thoughts:
First, the force of friction between the slab and the block,
Fs(max) = (mue)FN, and
FN = mg.
So,
Fs(max) = (mue)(mg)
Now, noting this relation, it makes sense to me that Fs(max) cannot be greater than FN... but this could also be wrong, because it's not that it can't be greater than FN, it just can't be greater than FN times (mue)... which is unknown... that's where I lose myself.
I just don't know what I need to consider, that I'm obviously not...
;)