Maximum horizontal force F on ramp with friction?

[PhoniexGuy]

Homework Statement

Assume the ramp and block make the following coefficients of friction: (μ_s=0.9 μ_k=0.3)
and the 2 kg mass is placed at rest on the ramp and a horizontal force F is also applied.

http://img69.imageshack.us/img69/4097/filepq.jpg

What is the biggest magnitude force F that can be applied and the mass remains stationary?

Homework Equations

I know that F_N = 20cos40 = 15.32N
F_|| = 20sin40= 12.85N

However will I need to subtract any components from Fn and so on?

The Attempt at a Solution

No idea what to do, I did the problem without friction and for that F has to be 16.8N, but no idea what to do with friction incorporated into it.

 

[ap123]

Free-body diagram?
Write expressions for the x and y components of all the forces.

The largest value of F corresponds to the largest value for the static frictional force.

If you can't solve it, show us your expressions for the components.

 

[TSny]

The Attempt at a Solution

No idea what to do, I did the problem without friction and for that F has to be 16.8N, but no idea what to do with friction incorporated into it.

Start with a carefully drawn free-body diagram for the block with all forces displayed and labeled. Decide on the orientation of your coordinate system - a good choice would be x-axis up along the slope and y-axis perpendicular to the slope. Use the diagram to find expressions for the x and y components of all the forces.

[I see ap123 beat me. Good. Note that we both start with FREE BODY DIAGRAM! ]

 

[haruspex]

I know that F_N = 20cos40 = 15.32N

No, it will be increased because of the applied force F.
List all the forces acting on the block, noting their directions, then write out the horizontal and vertical static equilibrium equations. Do the same for the ramp.

 

[PhoniexGuy]

Hmm, I think I get it. So acceleration is the F|| - ukFn right?
Then a = m(sin 40 - .3 * cos 40)
And i get acceleration =8.39 m/s/s
Then force F = 8.39/cos40 = 10.95N, so max F = 10.95 N? Or is this wrong?

 

[TSny]

Hmm, I think I get it. So acceleration is the F|| - ukFn right?
Then a = m(sin 40 - .3 * cos 40)
And i get acceleration =8.39 m/s/s
Then force F = 8.39/cos40 = 10.95N, so max F = 10.95 N? Or is this wrong?

Acceleration? The problem states that the block remains stationary.
I suspect that you still haven't drawn a free-body diagram. :grumpy:

 

[ap123]

What's happening here is that the block is stationary to start with.
As you increase the force, then eventually the block will start to move.
You need to calculate the value for F just before it starts moving.
This happens when the static frictional force reaches its maximum value.

 

[PhoniexGuy]

I found the acceleration the block would have if there was no horizontal force, so basically the force down the incline. And then to balance it i divided by cos 40. I drew the diagram, but still don't get it..

 

[PhoniexGuy]

@ap123, so the block is not moving in the start. So I just have to find the max friction can be, and then I have my component of max force F?

 

[ap123]

In the x-direction (along the ramp) you have the x-comp of F, the x-comp of the weight and the static frictional force. The sum of these will equal zero.
You will also need the normal force.
In the y-direction you have the y-comp of F and the y-comp of the weight and well as the normal force.
If you put all these together you will get the answer :)

Once you get all the components straight, it will be easier.

 

[haruspex]

So I just have to find the max friction can be, and then I have my component of max force F?

Yes.

 

[PhoniexGuy]

So, I think I got it.

Fn = mg cos theta + F sin theta
and then F = mg tan theta

So F = 16.9 N

and Fn = 26.10N, so max Ffric = 23.49 / cos 40 deg (component) = 30.7N for Max force F?

 

[ap123]

Nearly there!

Your first expression (for the y-components) is fine.
The second expression ( for the x-components ) needs to include the frictional force :
static friction + wsinθ - Fcosθ = 0

You then need to combine both equations - the only unknown will be F which you can solve for.

 

[PhoniexGuy]

Wait, what? So this is okay?
Fn = mg cos theta + F sin theta
And how do I set up the second equation? I don't understand what you wrote for it...

 

[ap123]

friction force down the ramp + x-comp of weight down the ramp = x-comp of F up the ramp

 

[PhoniexGuy]

So, does it become something like: Fn * Us + mg sin theta = F cos theta?

 

[ap123]

Looks good.

Take the value for the normal from the first equation and substitute it into this one.
Then solve for F and put in all the values.

Hope it works out!

 

[PhoniexGuy]

awesome, thanks!

 

[PhoniexGuy]

Hmm, I got -25N for the force of F? Is that right? Also can you explain the principle behind the second equation?

 

[haruspex]

Hmm, I got -25N for the force of F? Is that right? Also can you explain the principle behind the second equation?

No, it should be positive.
Your equations are F_n μ_s + mg sin(θ) = F cos(θ), and F_n = mg cos(θ) + F sin(θ), yes? If you can't find your mistake, please post your working from there.

 

[PhoniexGuy]

Hmm, ah, just misplaced negative, also, what is the purpose behind the second equation, why are you balancing those out?

 

[haruspex]

what is the purpose behind the second equation, why are you balancing those out?

The second equation expresses the fact that the block neither penetrates the ramp nor leaps off it. Without it (or something equivalent) you are not going to be able to solve it, are you?

 

[PhoniexGuy]

Hmm.. So it's keeping it static in the x direction? or parallel to the ramp?

 

[haruspex]

Hmm.. So it's keeping it static in the x direction? or parallel to the ramp?

You obtained F_n μ_s + mg sin(θ) = F cos(θ) by resolving parallel to the ramp. To complete use of the knowledge that the block does not move, you need to do likewise in at least one other direction. It doesn't matter which direction, so long as it is in the plane of interest. But an obvious choice would be normal to the ramp. This gives F_n = mg cos(θ) + F sin(θ) (which I believe you had earlier).

 

[PhoniexGuy]

Hmm, okay, that makes sense! Thanks, but is the the answer of 25.04 right? Wait, why is it plus and not minus? Fn μs + mg sin(θ) = F cos(θ)?