Sliding Block

[Zhalfirin88]

1. The problem statement, all variables and given/known data
A block with mass m = 5.0 kg slides down an inclined plane of slope angle 36.5o with a constant velocity. It is then projected up the same plane with an initial speed 1.05 m/s. How far up the incline will the block move before coming to rest?

3. The attempt at a solution

a_x = -g sin \theta

\Delta v^2 = 2a \Delta x

\frac{-v_o^2}{2a} = \Delta x

\frac{-1.05^2}{-2gsin \theta} = \Delta x

\frac{-1.05^2}{-11.658} = \Delta x

But this wasn't right.

[Doc Al]

As it slides down it's moving at constant velocity. So what force must be acting on the block besides gravity?

[Zhalfirin88]

I would say friction but there is no mention of it in the problem.

[Doc Al]

I would say friction but there is no mention of it in the problem.
Good. They expect you to conclude that on your own, based on the fact that it's not accelerating as it slides down.

So what must that friction force equal?

[Zhalfirin88]

Good. They expect you to conclude that on your own, based on the fact that it's not accelerating as it slides down.

So what must that friction force equal?

I'm getting

f = -mg sin \theta

[Doc Al]

I'm getting

f = -mg sin \theta
Good. The magnitude is mgsinθ, but the direction depends on which way the block moves.

[Zhalfirin88]

Good. The magnitude is mgsinθ, but the direction depends on which way the block moves.

I don't get how the friction helps any.

[Doc Al]

I don't get how the friction helps any.
To determine the acceleration, you need the net force. Friction is one of the forces acting on the block.

[Zhalfirin88]

You'd only do that when the block is projected back up the incline right? Then use the kinematic equation I had in OP?

Edit: So I was thinking about this more. When the block is projected back up the incline, you'd have friction and gravity in the X direction. Since I found the magnitude of the friction above, the acceleration would be:

+mg sin \theta + mg sin \theta = ma_x masses would cancel, so:

2 g sin \theta = a_x

Does this look right so far? (I set the positive X direction going down the incline)

Edit 2: Yes, I was right, thanks Doc :P

[Doc Al]

You'd only do that when the block is projected back up the incline right?
Friction acts whether the block is sliding up or down the incline. The problem you need to solve involves the block sliding up, so you need to find the block's acceleration as it slides up.
Then use the kinematic equation I had in OP?
Yes.

[Zhalfirin88]

Okay I have a new question that is similar to this.

A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is μs= 0.44, and the coefficient of kinetic friction is μk = 0.17. The angle of elevation θ is increased slowly from the horizontal. At what value of θ does the block begin to slide (in degrees)?

So I started out with:
Y-Direction
F_N - mg cos \theta = ma_y

F_N = mg cos \theta

X-Direction

-f + mg sin \theta = ma_x

-(\mu_s mg cos \theta) + mg sin \theta = ma_x

-\mu_s g cos \theta + g sin \theta = a_x

g(-\mu_s cos \theta + sin \theta) = a_x

But here is where I'm stuck and dunno what to do next.

[Doc Al]

Two hints:
- What does ax equal?
- When does static friction = μN?

[Zhalfirin88]

- When does static friction = μN?

I don't understand what you're asking, when there's no sliding?


- What does ax equal?

Since you're giving this as a hint, I'm guessing you mean it's equal to zero. But I thought that since the block is moving (question asks for kinetic friction) you don't know if ax is zero or not.

[Doc Al]

I don't understand what you're asking, when there's no sliding?
No, my point was that static friction does not always equal μN. μN represents the maximum value of static friction between two surfaces--the actual friction may well be less than that, depending on the situation. (For example: A book rests on a table. What's the static friction acting on it?)

Since you're giving this as a hint, I'm guessing you mean it's equal to zero.
Right.
But I thought that since the block is moving (question asks for kinetic friction) you don't know if ax is zero or not.
As the surface is tilted up, the acceleration remains zero until you get to the point where it just begins to start moving. That's the point you want, so it's only static friction that you care about. What's the maximum angle you can have and still not have the block slide? (Use the first hint, of course.)

[Zhalfirin88]

Yes, I knew that about static friction, however I think I'm stuck more mathematically, because if you simplify what I had up above when ax = 0 you get:

-\mu_s cos \theta + sin \theta = \frac{1}{g}
What next?

[Doc Al]

...if you simplify what I had up above when ax = 0 you get:

-\mu_s cos \theta + sin \theta = \frac{1}{g}

That's not quite correct. Take another look.

[Zhalfirin88]

Oops. I still don't see how to solve for the angle.



-\mu_s cos \theta + sin \theta = 0

[Doc Al]

That's better. Play with it a bit. Move terms around.

[Zhalfirin88]

Just a question, is this going to end up being an identity, because I don't know those very well. It'd help if I knew there was an identity involved.

[Doc Al]

No fancy trig identities involved. Hint: What trig function can be defined in terms of sine and cosine?

[Zhalfirin88]

Was in class. So:

\theta = tan^{-1} \mu_s

Now, is this always true? Like, if you're given a friction problem, and you need to find the angle, you can do this? Or vice versa, say you're given the angle and are asked to find the coefficient of static friction, you can immediately do this?

[Doc Al]

Was in class. So:

\theta = tan^{-1} \mu_s
Right.

Now, is this always true? Like, if you're given a friction problem, and you need to find the angle, you can do this? Or vice versa, say you're given the angle and are asked to find the coefficient of static friction, you can immediately do this?
What you found is the angle at which something just starts to slip as a function of the coefficient of friction. That's a specific answer to a specific question. Don't go blindly using tanθ = μ, just because a problem has friction and an incline.

Note that equations work two ways. Given the angle at which slipping starts, you can find the coefficient of friction.

[Zhalfirin88]

What you found is the angle at which something just starts to slip as a function of the coefficient of friction. That's a specific answer to a specific question. Don't go blindly using tanθ = μ, just because a problem has friction and an incline.

Note that equations work two ways. Given the angle at which slipping starts, you can find the coefficient of friction.

Hm, I didn't say what I wanted to say. What I wanted to ask is, when a mass starts slipping down an incline, will this always work? Or, will it only work when the angle of the incline is being increased?

[Doc Al]

What I wanted to ask is, when a mass starts slipping down an incline, will this always work? Or, will it only work when the angle of the incline is being increased?
If you mean anytime you have a mass starting to slip down an incline does tanθ = μ? No, not at all.