Need some help finding applied force & coefficient of static friction

[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

 

[Nathanael]

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

 

[summerchambers]

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

 

[Nathanael]

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

 

[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?

 

[Nathanael]

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

 

[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

 

[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 "[itex]\mu _s[/itex]"

and let's call the force that it is pressed against the surface with "[itex]F_{press}[/itex]"The coefficient of static friction is defined as follows:
[itex]\mu _sF_{press}=F_{max}[/itex]

This means that:
[itex]\mu _s=\frac{F_{max}}{F_{press}}[/itex]
In your case, [itex]F_{press}=mg=9.8m[/itex]
(this is because nothing is pushing down on the block other than it's own weight)

 

[summerchambers]

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

 

[Nathanael]

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

Yes, that is correct.

 

[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

 

[Nathanael]

but would it change if instead it was a smooth surface ? like glass

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.

 

[summerchambers]

ok thank you so much!