Object moving on a plane (frictionless)

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Homework Statement


A 2.50kg object is moving on a plane with coordinates x=2t2-4 and y=3t3-3.
Find the magnitude of the net force at t=2.00s

Homework Equations


[tex]\Sigma[/tex]Fy = 0 (no movement on y axis)
[tex]\Sigma[/tex]Fx = m*a (mass x acceleration)

The Attempt at a Solution


Alright, first of all, nice to meet you all; this is my first post on the forum so by all means, if I'm doing something incorrectly, just let me know.

Conceptualize: Using the sum of the froces acting on x and the sum of the forces acting on y, I think I could find the magnitude of the net force with
|F|2= x2+y2

1- I figured it would be easier to draw the diagram of the forces by taking (x,y) coordinates at t=0 & t=2 (in order to see the direction of the incline). This one seems to be going from the 3rd quadran to the first.

2- I can then write my [tex]\Sigma[/tex]Fx & [tex]\Sigma[/tex]Fy equations like this:

[tex]\Sigma[/tex]Fy = 0 = n - w
n = normal force
w = weight = mass x gravity(acceleration)

[tex]\Sigma[/tex]Fx = m * (F sin [tex]\theta[/tex])

This is where I get confused.
Should I include a y component for the force in my y (F cos [tex]\theta[/tex])?
And what other forces are acting on my x axis??

P.S.: I seem to be having problems with the code @ point 2-; if you see a square root in front of Fx, ignore it, it's a SIGMA... I tried refreshing the page and all but I keep getting the sqr root in the preview ô.O
Same goes for F sin theta... it's obviously not F sin sigma...
 

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Answers and Replies

  • #2
kuruman
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Hi Kumos, welcome to PF. I read the problem as saying that the x and y coordinates of the object have the time dependence that you posted. Knowing this, can you find the x and y components of the acceleration?
 
  • #3
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I'm sorry I forgot to mention t was in seconds so yes, you're right about the time dependance.
Yeah I thought about acceleration but...
Acceleration = displacement/time
ax = x2 -x1 / t2 -t1
ax = 4 - (-4) / 2 -0 = 4 m/s2
ay = y2 - y1 / t2 - t1
ay = 21 - (-3) / 2-0 = 12 m/s2

But once I do have my acceleration, if I use the equation for an object in equilibrium (y-axis for example) I get:
[tex]\Sigma[/tex]Fy = m*a = 0
[tex]\Sigma[/tex]Fy = 2.50kg * 12m/s2 = 30 N
Which can't be since the object is in equilibrium on the y axis the sum of the forces have to equal zero right? Or am I wrong in my m*a maybe?? I'm thinking 'a' is off...

I was thinking of finding all the components of the force on x-axis and y-axis using theta.
And then plugging in my numbers.
Following that idea (see my graph attached to 1st post for the variables)
[tex]\Sigma[/tex]Fy = 0 = n + Fsin[tex]\theta[/tex] - m*g
Where g=gravity & n=normal force
[tex]\Sigma[/tex]Fx = Fcos[tex]\theta[/tex] = m*a

I know : m, g & theta (using tan-1)
I don't know: n, F & a
So can I use ax= 4m/s2? Since ay doesn't sound right, using ax sounds sketchy...
I think I might be missing an equation...

As you can see, I'm really confused. :/
Am I on the right train of thoughts at least?
 
  • #4
kuruman
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Acceleration = displacement/time
ax = x2 -x1 / t2 -t1
ax = 4 - (-4) / 2 -0 = 4 m/s2
Absolutely not.

Try the definitions

[tex]v_x=\frac{dx}{dt}[/tex]

[tex]a_x=\frac{dv_x}{dt}[/tex]

and similar expressions for the y-direction.
 
  • #5
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Thanks for your answer but using the derivatives I find
ay = 18t m/s2
And at t = 2.00s I get a net force of 90N on the y-axis (m*a).

I guess my main question still hasn't been adressed ([tex]\Sigma[/tex]Fy & [tex]\Sigma[/tex]Fx components) and I'm spending more time posting here then adressing the problem.

Thanks anyways.
 
  • #6
kuruman
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Thanks for your answer but using the derivatives I find
ay = 18t m/s2
And at t = 2.00s I get a net force of 90N on the y-axis (m*a).
Correct. You can now use Newton's Second Law to find Fy.

What about the acceleration in the x-direction? Can you find it with the same procedure? If so, then you can find Fx.

I guess my main question still hasn't been adressed ([tex]\Sigma[/tex]Fy & [tex]\Sigma[/tex]Fx components)
You know that ΣFy = may. You already know ay and m, therefore you can find ΣFy. If you find ax, you can calculate ΣFx the same way. So why is the question not addressed? You seem to think that the object is in equilibrium. It cannot be because it is accelerating.
and I'm spending more time posting here then adressing the problem.
Are you saying that posting here while someone helps you become unconfused is not worth as much as looking at the problem and remaining confused? I know it isn't so and I understand your frustration, but comprehension has to come from within.
Thanks anyways.
You are welcome.
 
  • #7
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You know that ΣFy = may. You already know ay and m, therefore you can find ΣFy.
Isn't that
And at t = 2.00s I get a net force of 90N on the y-axis (m*a).

You seem to think that the object is in equilibrium. It cannot be because it is accelerating.
Having set my y-axis to be perpendicular to the incline plane, yes I did think it was in equilibrium on that axis. It's not the case?
Having an incline y-axis makes the object move solely on the x-axis no?

and I'm spending more time posting here then adressing the problem.
Are you saying that posting here while someone helps you become unconfused is not worth as much as looking at the problem and remaining confused?
No, I'm referring to me spending 20min writting a post and still not understang the conceptualization of the problem. But if indeed the object is in motion on the y-axis, things became a lot clearer. :wink:

So then ax = 4m/s2
@ t=2s
ΣFy = 90N
ΣFx = 10N

Find the magnitude of the net force at t=2.00s
|F|net = sqr 102 + 902 = 90.6N ?
And the answer went through.

I believe I was looking way too much into this... (see my graph)
I'll also work on making my posts shorter :wink:
 
  • #8
kuruman
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You made the assumption that the object is on an inclined plane and that gravity acts on the object. I don't think this is the case here. The object is on a horizontal plane and a single force acts on it so that its position as a function of time is what you posted.

Yes, you were seeing too much into it, but I would say don't make any heroic effort to make your future posts shorter. It is easier to help someone if there is a lot of information about what is on his/her mind. :smile:
 
  • #9
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Thanks again for everything!
If the coordinates are the ones of the displacement of the object and not the ones of the force... isn't it better to modelize the object on an incline plane (since there are both x,y components)?
Is it that regardless of the plane, the displacement of the object can be assumed to undergo the same x,y components as the ones of the Force and thus we can use these components as the force components? That actually makes sense I think...

Let me ask you another question, I was unsure wether I should create a new post for this so just let me know if that is the case.
In the Atwood Machine, the theory we were given to find an expression for ay and T(tension) is:
1) [tex]\Sigma[/tex]Fy1= T - m1g = m1ay
2) [tex]\Sigma[/tex]Fy2 = m2g - T =m2ay
3) ''Add equation (2) and equation (1) noticing that T cancels''
4) Solve for ay
5) Solve for T using either of equation (1) or (2)

My question is... why do we add equation (1) & (2) at the 3rd step??
Because we're looking for the total acceleration of the system (ay)?
What if the system is not moving (equilibrium)... ay = 0; so can we postulate that equation (1) = equation (2)?
 
  • #10
kuruman
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Thanks again for everything!
If the coordinates are the ones of the displacement of the object and not the ones of the force... isn't it better to modelize the object on an incline plane (since there are both x,y components)?
No. Once you are given the position vector as a function of time, the acceleration vector is uniquely defined as the second derivative of the position. It is what it is. In this particular case, the y-component of the acceleration depends on time. For an inclined plane the acceleration is constant, so the inclined plane is a poor model for this particular case.

Let me ask you another question, I was unsure wether I should create a new post for this so just let me know if that is the case.
You should have created a new thread, but OK for this time.
In the Atwood Machine, the theory we were given to find an expression for ay and T(tension) is:
1) [tex]\Sigma[/tex]Fy1= T - m1g = m1ay
2) [tex]\Sigma[/tex]Fy2 = m2g - T =m2ay
3) ''Add equation (2) and equation (1) noticing that T cancels''
4) Solve for ay
5) Solve for T using either of equation (1) or (2)

My question is... why do we add equation (1) & (2) at the 3rd step??
Because we're looking for the total acceleration of the system (ay)?
Yes, that is the reason.
What if the system is not moving (equilibrium)... ay = 0; so can we postulate that equation (1) = equation (2)?
Yes, again, in which case you have to conclude that T = m1g = m2g which in turn implies that the masses have to be equal.
 
  • #11
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Thank you :)
 

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