# Need some help finding applied force & coefficient of static friction

1. Jul 25, 2014

### summerchambers

i am having some trouble figuring this out

I can not figure out which formula to use i have tried Mass x G x Static Friction
which resulted in 1 x 2 x 9.8 = 19.6

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2. Jul 25, 2014

### Nathanael

Mass * g * coeffecient of static friction = "Maximum value of static friction"

3. Jul 25, 2014

### summerchambers

when i had the spring it doesn't give me the coefficient it just gives me the Fmax

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4. Jul 25, 2014

### Nathanael

But you can still use Fmax to find the coefficient. (Rearrange the equation)

5. Jul 25, 2014

### summerchambers

''Mass * g * coeffecient of static friction = "Maximum value of static friction"

so it would be like mass * g / maximum value of static friction = coefficient of static friction?

6. Jul 25, 2014

### Nathanael

No your algebra is a little wrong. You would want to divide both sides by (mass * g)

7. Jul 25, 2014

### summerchambers

i don't really understand , can you please give me an example ? I think i get what you mean but i think I'm wrong as well

8. Jul 25, 2014

### Nathanael

I think your getting confused because we're using words instead of letters.

But let's start from the beginning. What does static friction depend on? Well it obviously depends on the two surfaces (a block on a smooth surface will have less friction than a block on a rough surface) but it also depends on how strongly the block is "pressed against" the surface.
(If you don't believe me, try pushing something across a table while pushing down on it very hard.)

So what is the force that the block is "pressed against" the surface with? It's just the weight of the block (which is mass * g)

Now,
Let's call the coefficient of static friction "$\mu _s$"

and let's call the force that it is pressed against the surface with "$F_{press}$"

The coefficient of static friction is defined as follows:
$\mu _sF_{press}=F_{max}$

This means that:
$\mu _s=\frac{F_{max}}{F_{press}}$

In your case, $F_{press}=mg=9.8m$
(this is because nothing is pushing down on the block other than it's own weight)

9. Jul 25, 2014

### summerchambers

so it would be like µs = 2/9.8
for the first trial ?

10. Jul 25, 2014

### Nathanael

Yes, that is correct.

11. Jul 25, 2014

### summerchambers

and so on and so forth for the rest of the trials ? but would it change if instead it was a smooth surface ? like glass

12. Jul 25, 2014

### Nathanael

If it were a smooth surface (or any different surface) then the coefficient of static friction would change

But the method of finding the coefficient of static friction would still be the same.

13. Jul 25, 2014

### summerchambers

ok thank you so much!!