# An incline plane, friction, and a pulley.

1. Aug 16, 2007

### Idioticsmartie

1. Q: With what amount of force must a person pull on the vertical portion of the rope to make the 1000 kg box travel up the 45 degree rough ($$u_{k}$$ = 0.10) plane at constant speed?

Umm, here is a really awful diagram to help you get the picture. Mac doesn't come with a Paint program, so I grabbed a random one...and this is what I got. Again, sorry.

$$u_{k}$$ = 0.10
m = 1000 kg
$$\vartheta$$ = 45
g = 9.80 $$m/s^{2}$$

2. Relevant equations
$$\Sigma$$F = m*a
$$F_{fr}$$=$$u_{k}$$*$$F_{n}$$

3. The attempt at a solution

Known Forces:
mg = 9800
$$F_{n}$$ = 6929.646
$$F_{fr}$$ = 692.9646

I think I did this problem correctly, but would someone mind terribly checking it? It's just that there aren't any examples in my book, so I want a second opinion before I turn this in to my teacher. Thanks!

1) The co-ordinate plane is set so that the x-axis is the inclined plane and the y-axis is perpendicular (of course), so $$F_{n}$$ = y-axis.
$$F_{n}$$ is set perpendicular to the plane, while mg goes straight down. When you resolve mg, you get <6929.646, -6929.646>.

2) Therefore, $$F_{n}$$ = the y-component of mg, so $$F_{n}$$ = <0,6929.646>

3) $$F_{fr}$$ = $$u_{k}$$$$F_{n}$$ = 6929.646 * 0.10 = 692.9646

4)$$\Sigma$$ F = $$F_{a}$$- $$F_{fr}$$
$$F_{a}$$ > $$u_{k}$$ * $$F_{n}$$
$$F_{a}$$ = 693.
Sigfigs, so it's 690 kg.

Is this correct?

Last edited: Aug 16, 2007
2. Aug 16, 2007

### rootX

I don't think so.
yea, you missed one force(as radou said) lol

Last edited: Aug 16, 2007
3. Aug 16, 2007

Didn't you miss a force?

4. Aug 16, 2007

### Idioticsmartie

What am I missing?

Sorry about the formatting issues - I'm just beginning to get the hang of Latex

5. Aug 16, 2007

### rootX

You got that 6929.646, but then you never used it in your force equation.

Shouldn't it be like <-6929.646, 6929.6>?

Last edited: Aug 16, 2007
6. Aug 16, 2007

### Idioticsmartie

But it's <x,y> and since the x-axes is parallel to the inclined plane, the yis negative.

Sorry, I'm really slow at physics. But I don't quite understand why you need to factor in mg to the equation. The applied force has to be greater than the friction force in order for the box to move, correct? And the friction force is u$$_{k}$$ * F$$_{n}$$

So where does the mg fit in?

Again, sorry.

7. Aug 16, 2007

### rootX

yea, you are right, I skipped one step, and straight way started thinking aout N.

As, it's a slope so, a part of gravity (that is parallel to the x-axis) also needs to be considered. So, you should have three horizontal forces(a part of the gravity, friction, and applied force) acting on the object.
[I am really poor at explaining things ><]
Hopefully, this helps you :D

8. Aug 16, 2007

### Idioticsmartie

Okay, so let me get this straight:

The sum of the forces on the object are:
The applied force (which is, as of yet, a mystery)
Friction
And the x-component of gravity. (Right? I need to verify this)

So:
Seeing as the applied force has to be greater than the friction force, do I subtract the x-component from the applied force (693 N)?

Wait; the gravity is working against the applied force, isn't it? So I have to add it to the applied force in order to move the box? So that would be 693 + 6929, wouldn't it?

Or am I totally off? Thanks a lot for helping with this!

9. Aug 16, 2007

### Gear300

heheheheh....I lul'd at the picture

10. Aug 16, 2007

### Idioticsmartie

in my defense, there was no way to draw a straight line with the darn tool.

I'm totally a mac over windows person, but in this case, I kinda vote for Paint.

11. Aug 16, 2007

### Gear300

To move the box up at constant speed means 0 acceleration, so the force you apply vertically (an external force) should cancel out the the forces acting on the box within the system, meaning what is moving the box other than the vertical pull.
No, gravity is the force moving the load down the plane. To find the the force its moving down the plane, you use: Fgx = mgsin(theta). To find the friction, you find the normal force, which is Fn = Fgy = mgcos(theta). Once you have normal force, you multiply it by the constant .10 to get the friction. Once you have this, you subtract the friction force from the force moving the box, which is Fgx...and so, you know what to do from there.

12. Aug 16, 2007

### Gear300

I prefer Windows over Mac...just that when I used Mac, I wasn't well adjusted with it.

13. Aug 16, 2007

### Idioticsmartie

I'm afraid that your explanation only served to confuse me more. Why is it that the force moving the box up the plane is Fgx and not simply more than the friction force?

14. Aug 16, 2007

### rootX

yep

yea, so
Applied force-Weight/sqrt(2)-frictional force = 0

it's sqrt(2) because only horizontal forces are being taken into account.

15. Aug 16, 2007

### Idioticsmartie

Okay, so:

applied force = sq.rt of weight + frictional force?

So that's 31.6227766 + 692.646 = 724.571

So if we account for significant figures, I get a grand, final answer *dum de dum* of 720 N?

Why does it equal zero?

Last edited: Aug 16, 2007
16. Aug 16, 2007

### rootX

no!
applied force = horizontal comp. of W + frictional force
=7622 N

hmm.. hopefully this is correct ><

17. Aug 17, 2007

### Idioticsmartie

so then where did the sq. rt part come from?

18. Aug 17, 2007

### rootX

cos/sin (45) = 1/sqrt(2)

19. Aug 17, 2007

### Idioticsmartie

Alright, so with 2 sigfigs, it's 7600 N. :)

Thanks so much for the help- I would've done badly on the homework otherwise!