Calculating Force and Motion on an Inclined Plane

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Femme_physics
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Homework Statement

http://img20.imageshack.us/img20/4519/objectssss.jpg

A body whose weight is 12 [kN] progresses upwards on an inclined plane, slanted to the horizontal line at angle alpha. Force F, acting on the body, parallel to the inclined plane, remains constant during the entire action. During the body progression at segment AB (whose length is L) its velocity decreases from Vo to V. The kinetic friction between the object and the surface is u.

A) Calculate force F
B) If force F won't act on the object, will the object stay at rest or will it slide down the inclined plane? Presume that the static and kinematic friction coeffecients are equal.

The Attempt at a Solution



Did I get it right so far? I'm not sure where do I input velocity. I do know there is no movement on the y-axis (with the coordinates I've chosen) but only on the x axis. So it made sense to me that sigma Fy = 0 and sigma Fx = ma.

What do you say?

http://img864.imageshack.us/img864/145/90689814.jpg http://img853.imageshack.us/img853/989/66580583.jpg
 
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ehild said:
What is the question?

ehild

A) Calculate how stupid is Femme_Physics for not including the question?
B) Calculate how hard Femme_Physics slapped her forehead for that?...


...


The questions really are:

A) Calculate force F
B) If force F won't act on the object, will the object stay at rest or will it slide down the inclined plane? Presume that the static and kinematic friction coeffecients are equal.
 
Femme_physics said:
A) Calculate how stupid is Femme_Physics for not including the question?
B) Calculate how hard Femme_Physics slapped her forehead for that?...

Oh don't do it! It will damage your brain and it would be catastrophic ... :wink:

ehild
 
Hmm. So "a" is equal delta v divided by L?

I just wrote the sum of all forces = ma.

Is my sum of all forces correct, at least?

Oh don't do it! It will damage your brain and it would be catastrophic ...

::blushing::shy:

I'm applying just enough Newtons to jolt not to permanently damage :approve:
 
Given what you have, consider using
[itex]\Delta d = \frac{v_{2}^{2}-v_{1}^{2}}{2a}[/itex]
to find the acceleration. Your force sum looks fine, and that should give the unknown force.
 
Femme_physics said:
Hmm. So "a" is equal delta v divided by L?

Have you slapped your head already? :smile: What is the dimension of acceleration, is it the same as that of (delta v)/L?
You have a simple one dimensional motion with constant acceleration.

ehild
 
Femme_physics said:
So I used the wrong formula. Huh.

Alright, well how about this for F? Looks good?

*crosses fingers* please say it's all correct please say it's all correct

*Looking at your fingers*
Yes, your fingers look nice! :smile:

And yes, your a and F are correct too! :wink:

Btw, I would write L instead of Δd.
And I would write v0 and v instead of v1 and v2.
And actually, I find F = 7133.6 [N], that is, without the rounding errors you made.
 
I like Serena said:
*Looking at your fingers*
Yes, your fingers look nice! :smile:

And yes, your a and F are correct too! :wink:

Btw, I would write L instead of Δd.
And I would write v0 and v instead of v1 and v2.
And actually, I find F = 7133.6 [N], that is, without the rounding errors you made.
I beg your pardon good sir! I made no such "rounding errors". I may have overrounded, but I did not erroneously round. To do such would mean that I rounded 1.7 to 1. That is a rounding error. Overrounding is to round 1.7 to 2 instead of leaving it 1.7.

What about question B?
Question B seems like a basic statics question. I'll tackle that knuckle tomorrow^^

Thanks!
 
Femme_physics said:
I beg your pardon good sir! I made no such "rounding errors". I may have overrounded, but I did not erroneously round. To do such would mean that I rounded 1.7 to 1. That is a rounding error. Overrounding is to round 1.7 to 2 instead of leaving it 1.7.

I pardon you good madam! I may have been overzealous with rounding. (Forgive me for being a nitpicker. ;))
The proper answer is actually F = 7 kN.
Yes, that is only 1 significant digit! Because really, that is al you have! :smile:


Femme_physics said:
Question B seems like a basic statics question. I'll tackle that knuckle tomorrow^^

Thanks!

See you tomorrow then? Because then I will have a bit more time than today and I would like that! :smile:
 
FP, check the text of the problem. "During the body progression at segment AB (whose length is L) its velocity decreases from Vo to V." Vo=6 m/s, V=3 m/s. Is the acceleration positive then?
All of us overlooked this mistake.ehild
 
(Just woke up. :zzz:)

ehild said:
FP, check the text of the problem. "During the body progression at segment AB (whose length is L) its velocity decreases from Vo to V." Vo=6 m/s, V=3 m/s. Is the acceleration positive then?
All of us overlooked this mistake.

ehild

Yes, you're right.
 
Ahh...so I just switch the 3 and the 6. Let me get to it :) And then question B looks like statics. I'll be cracking it now! Thanks! :)
 
I'm looking for the acceleration formula we were using.. this is it right?

theformula.jpg
 
Femme_physics said:
I get the same result for a just in minus. I see why now, since the object is decelerating, it makes sense.

Sorry, there is still more to it, since the result for F will be different now...


Femme_physics said:
Statics was a cakewalk ;)

Thanks, you two! XoX

Very good! :smile:
 
I know there's more to do but I solved it using a pencil and I'd really hate to start recopying it with black marker. I just changed 0.45 to -0.45, used that as acceleration and got my answer :)

Here...see if you can see anything

http://img815.imageshack.us/img815/3388/ifucansee.jpg
 
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Femme_physics said:
I know there's more to do but I solved it using a pencil and I'd really hate to start recopying it with black marker. I just changed 0.45 to -0.45, used that as acceleration and got my answer :)

Here...see if you can see anything

*looking*
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*

Yep! I have the exact same answer! :smile:

Aw, this was over all too quickly! Do you have another cookie? :wink:
 
I like Serena said:
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*
*looking*

Yep! I have the exact same answer! :smile:

Aw, this was over all too quickly! Do you have another cookie? :wink:

w00t ;)
*chuckles* Yes am posting it right now, :) I just didn't want two dynamics problems to clash! That'd create a hole in the space time continuum, and all.
 
Actually, as I'm writing the problem, I'm coming up with ideas so that's why I still haven't posted a new problem :) It's amazing what just typing the problem or reading it out loud can do.
 
Femme_physics said:
Actually, as I'm writing the problem, I'm coming up with ideas so that's why I still haven't posted a new problem :) It's amazing what just typing the problem or reading it out loud can do.

Indeed. That's why usually people work in groups :)
 
Good morning, ILS! :smile:

Am getting same result
 
It's a plus sign!