Inclined Plane Static Friction Coefficient

[Mark Trice]

Homework Statement

The static friction coefficient on an inclined plane is equal to tan(θ). The problem requires me to prove this also works by using an x-axis parallel to the ground rather than parallel to the incline. I understand how to get tan(θ) using the incline as the x-axis, but I don't know how to get tan(θ) using an x-axis parallel to the ground.

Homework Equations

μ = tan(θ)

The Attempt at a Solution

I worked out how to get tan(theta) using the incline as the x-axis by doing mgsin(θ) = (μ)(Normal Force) and mgcos(θ) = Normal Force and used algebraic manipulation to get μ = tan(θ), but I just can't wrap my head around how the trigonometry would work if I used a different x-axis.

 

[kuruman]

Hi Mark Trice and welcome to PF.

Will you please post the statement of the problem exactly as was given to you? Thanks.

 

[Mark Trice]

The problem states "μ = tanθ is true no matter the axis you choose on an inclined plane. Prove this by making the x-axis parallel to the ground and solving for μ to get tanθ".

 

[kuruman]

The problem states "μ = tanθ is true no matter the axis you choose on an inclined plane. Prove this by making the x-axis parallel to the ground and solving for μ to get tanθ".

Solve what for μ? What is the physical situation? There must be more to this. In any case, it sounds like you have to draw a different free body diagram (?) with rotated axes relative to one you already have. How about starting from there? Please post your work in detail.

On edit: Imagine putting a small box on a plywood sheet. The coefficient friction between the two is, say, μ = 0.3. Now lift one end of the plywood so that it forms angle θ with the horizontal. If you increase the angle, tanθ will increase. If μ = tanθ, then μ should also increase to a value larger than 0.3 and keep on increasing as you increase the angle. Does this make sense to you?

 

[Mark Trice]

Sorry for the confusing question; this question is based on a lab we did and I forgot to explain the situation of the lab. So basically in the lab we raised a piece of wood at an incline until a block on the inclined plane began to slide by overcoming the static friction. The angle θ is the angle that the incline makes with the ground. μ is the static friction coefficient on the inclined plane. So the question is dealing with a block that hasn't yet overcome the static friction, so it is sitting stationary on an inclined plane. I need to show that μ = tanθ is true of the block resting on the inclined plane by using an x-axis parallel to the ground (not an x-axis parallel to the incline!).

Let me know if I need to clear anything up.

 

[kuruman]

Sorry for the confusing question; this question is based on a lab we did and I forgot to explain the situation of the lab. So basically in the lab we raised a piece of wood at an incline until a block on the inclined plane began to slide by overcoming the static friction. The angle θ is the angle that the incline makes with the ground. μ is the static friction coefficient on the inclined plane. So the question is dealing with a block that hasn't yet overcome the static friction, so it is sitting stationary on an inclined plane. I need to show that μ = tanθ is true of the block resting on the inclined plane by using an x-axis parallel to the ground (not an x-axis parallel to the incline!).

Let me know if I need to clear anything up.

Thank you for providing the context. It is now clear what you have to do. Draw a free body diagram with the axes as required by the question. Note that the weight vector is entirely along the negative y-axis and the the force of friction and normal force, have x and y components. Write Newton's 2nd law twice, once for the new x-direction and once for the y-direction. Assume that the block is on the verge of slipping. It's a bit more complicated than the other coordinate system but doable if you are careful with your trig functions and your algebraic signs. If you think you need more help, please post your work. An uploaded photograph of your diagram might be sufficient if it's legible and has good contrast.

 

[Mark Trice]

Here is my attempt at this so far. I can barely even start this because doing this without rotating the axis just doesn’t make sense to me. Should I put my x-axis through the center of the object parallel to the ground rather than on the ground?

 

[Mark Trice]

Now I have a little bit more, but have no clue where to go now.

 

[haruspex]

Now I have a little bit more, but have no clue where to go now.

You have written that $F_{sx}=\mu_{sx}F_{Nx}$.
First, $\mu_{sx}$ doesn't mean anything. μ_s is just a number. It cannot have x and y components.
Second, the frictional force depends on the normal force, not on just some component of the normal force.
You know that the frictional force has magnitude μ_sN. You know the direction of that force. What is its x component?
Third, you have written that the x component if the normal force is F_Ncosθ. Check that. Does it make sense when θ=0?

 

[Mark Trice]

(μs)cosθ for the x-component?

Oh crap, would it instead be Fnsinθ?

edit: Sorry, not sure how to do subscripts on here

 

[Mark Trice]

So I managed to figure it out surprisingly. Thank you all for the help.

 

[haruspex]

not sure how to do subscripts on here

In the bar above the text entry area, look for X₂; and X² for superscript.