An incline plane, friction, and a pulley.

[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.

http://photos-a.ak.facebook.com/photos-ak-sctm/v123/24/75/1238100168/n1238100168_30084828_487.jpg

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

Homework Equations

\SigmaF = m*a
F_{fr}=u_{k}*F_{n}

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 into 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?

 

[rootX]

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.

http://photos-a.ak.facebook.com/photos-ak-sctm/v123/24/75/1238100168/n1238100168_30084828_487.jpg

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

Homework Equations

\SigmaF = m*a
F_{fr}=u_{k}*F_{n}

The Attempt at a Solution

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 into 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?

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

 

[radou]

Didn't you miss a force?

 

[Idioticsmartie]

What am I missing?

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

 

[rootX]

What am I missing?

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

When you resolve mg, you get <6929.646, -6929.646>.

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

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

 

[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.

 

[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

 

[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!

 

[Gear300]

heheheheh...I lul'd at the picture

 

[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.

 

[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.

 

[Gear300]

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

 

[Idioticsmartie]

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.

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?

 

[rootX]

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)
!

yep

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)?

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


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

 

[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?

yep
yea, so
Applied force-Weight/sqrt(2)-frictional force = 0it's sqrt(2) because only horizontal forces are being taken into account.

Why does it equal zero?

 

[rootX]

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

hmm.. hopefully this is correct ><

 

[Idioticsmartie]

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

 

[rootX]

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

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

 

[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!