Coefficient of Static Friction = tan (angle of incline)

[lem0ncheezcake]

Homework Statement

I am trying to prove that the coefficient of static friction is equal to the tan of the angle of incline. (You can find the proof of this from )

I set the angle of incline as my independent variable and had an angle range from 10 to 37.5 degrees. After setting the slope to different angles, I measured the extra force required to cause the wooden block to begin to move on the slope. I did this by connecting a string to the wooden block and to a container that could be filled with sand (using a pulley to connect them).

Homework Equations

μ = (mg sin(θ) + Mg)/(mg cos (θ))

where m is the mass of the wooden block and M is the mass of the handing container and sand.

This simplifies down to μ = tanθ + M/(m cosθ)

However, it is also known that μ = tanθ

Equating the two equations we get tanθ + M/(m cosθ) = tanθ, which is impossible. Can anyone explain what I've down wrong here?

The Attempt at a Solution

I tried manipulating the equation,

M/(m cosθ) = μ - tanθ

=> M = μmcosθ - msinθ

=> M = m(μcosθ - sinθ)

Ultimately, I aim to draw a graph which shows μ = tanθ, however, with the values I obtained so far, no such graph can be drawn.

I would really appreciate it if someone could help me!

 

[BvU]

Hello LCC,

Nice experiment ! Well described in this your first post, kudos !

to cause the wooden block to begin to move on the slope

Sounds like you filled until m started to move upwards along the slope. am I right ? If so, do $\mu m g \cos\theta$ and $mg\sin\theta$ point in opposite directions, as your equation suggests ?

 

[lem0ncheezcake]

Hello LCC,

Nice experiment ! Well described in this your first post, kudos !
Sounds like you filled until m started to move upwards along the slope. am I right ? If so, do $\mu m g \cos\theta$ and $mg\sin\theta$ point in opposite directions, as your equation suggests ?

Hi BvU,

Yes, you got the idea right, except I set it up so that m could start to move downwards along the slope.

I am not sure to be honest. I do know however, mgcosθ and mgsinθ are perpendicular to each other because they represent the vertical and horizontal components of force due to the weight of m. Does this help?

 

[BvU]

However, it is also known that μ = tanθ

This is for the situation as in the video when there is no extra force involved !

I set it up so that m could start to move downwards

Ok, so your μ = tanθ + M/(m cosθ) has the right sign and you have a set of observations of M as a function of $\theta$. You can investigate if $\mu$ depends on $\theta$ (*). But if you want to show that $\mu = \tan\theta$ directly, you'll have to find a way to vary $\mu$ and work with M = 0.

(*)
$\mu = \tan\theta$ doesn't mean that $\mu$ varies with $\theta$; it means that the angle at which sliding is about to start has a tangent with a value that is equal to $\mu$.

 

[Chestermiller]

If the pulley is at the top of the incline and you are adding mass M to try to pull the block up the incline, then the force balance on the block is
$$Mg-mg\sin{\theta} = F \leq mg\mu cos{\theta}$$where F is the friction force. So, the coefficient of friction satisfies the inequality
$$ \mu \geq \frac{M}{m}\sec{\theta}-\tan{\theta} $$and M satisfies the inequality:
$$M\leq m(\sin{\theta} + \mu cos{\theta})$$
The equal sign applies when the block is just on the verge of sliding.