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[Max White]

So assuming I had two coins, both identical in size and shape, but one is exactly twice the mass of the other. I put these both on a wooden plank until one or both starts to slide down the plank. What one will begin to slide first? Also I don't just need the answer, I need the equations involved, remembering that the force of friction is directly proportional to the normal force.
Thanks for any help, I only have a few hours to figure this out and it will take me longer than that just to find my old textbooks...

[Pnx]

Not that I can answer this, but for clarity are you saying something along the lines of the plank is gradually tilted by you until one of the coins begin to slide?

[Max White]

That is exactly what I'm saying.
For context, a friend of mine is doing firsr year sciene, but she is in a somewhat bad financial situation and can't afford a good scientific calculator. One of her lecturers challenged the class to find the solution on a few hours, with the best answer winning a really nice calculator.

[darklord92]

you need to calculate the force of friction between the objects

first we need the normal so:   

Normal = Mass* Gravity

you will need to find static co-efficient between wood and copper, after you have that plug the numbers in here:

"Static force to overcome in newtons" = coefficient *  Normal force


Once you have the lower number will be the coin to move first

[Arx]

The frictional force opposing a force on a static object is given by the formula
f = μ_s*N
where μ_s is the static friction coefficient, and N is the normal force.

The normal force is equal to the component of the object's weight perpendicular to the surface it's on; thus on a flat surface
N = m*g;

and on an incline
N = (m*g)sin or cosθ.
depending on the direction of the incline, I think.

Therefore, the coin with the lesser mass will slide first; more detail can't be given with that much information.

I would have responded sooner, but I was busy writing a test on almost exactly this topic...

Detail edited in!

[Jelle]

They will slide at the same time? I'm a bit rusty at this so I really don't know for sure.

Here's  me blurting out some math.

f1=fn+f+fr
 =cosyfg+sinyfg-xsinyfg
 =cosym1g+sinym1g-xsinym1g
f2=cosym2g+sinym2g-xsinym2g

Where all these forces are vectors of course but to lazy for propper notation.

Where g is the same naturally, so I made the equation.

m1a1/(cosym1+sinym1-xsinym1)=m2a2/(cosym2+sinym2-xsinym2)
a1/(cosy+siny-xsiny)=a2/(cosy+siny-xsiny)
a1=a2

Where fn normal force, f force down the surface of the board and fr the resistance force. m mass a acceleration, y the angle of the surface to horizontal, x the resistance factor of the material.

I'm probably wrong, like I say rusty. It's a gravitational force and a force directly proportional to it though, so given the same proportional resistance they should slide at the same time at the same acceleration.

Edit: I probably forget to note why I consider x1 and x2 the same. Since the coins are of the same size and material  I naturally took them to be the same. If they were not the same size the final step would give a1x2=a2x1 (or a1/x1=a2/x2 if you prefer).

[Max White]

We ended up drawing a triangle and figuring out the forces from there. Final equation was that it slides when tanθ > μ, no mass, therefor they slide at the same angle, but the important part was showing all the working.

[jhxmt]

My simple (non-physics-educated) mind went:

"Friction is the force between the coins and the wood, driven by their weight i.e. gravity and their mass.  So if one's got twice the mass, it'll have twice the friction.

But the force that causes them to slide at a certain angle is also driven by their weight i.e. gravity and their mass.  So if one's got twice the mass, it'll have twice the sliding force at any given angle.

So they'll slide at the same angle, at the same time."

Had absolutely no idea how that would be put into formulae.  This is why my physics education floundered somewhat.