Y directed force as a function of time

AI Thread Summary
The discussion revolves around calculating the y-directed force as a function of time for a plane with a mass of 300,000 kg, height of 10 miles, and initial speed of 100 mi/hr. The original force equation provided was questioned for accuracy, suggesting it might be off by a factor of ten. Participants emphasized the need for a clear problem statement and understanding of the equations involved, particularly distinguishing between horizontal and vertical forces. The correct calculation for the force was clarified to be approximately 4.8 x 10^9 N, and the importance of reviewing each step in the problem-solving process was highlighted. Overall, a thorough understanding of the problem's context and equations is essential for accurate calculations.
tj77
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Homework Statement
Assume the mass of the plane is 300,000kg h = 10 miles L = 25 miles and vxo = 100 mi/hr.
Determine the lift(or y-directed force not including the force of gravity) as a function of time.
Relevant Equations
https://imgur.com/a/43fivhe
The question before asked to find the net force as a function of time which I got:

F = 4.83×10¹⁰ (1.65×10⁻⁸ t − 7.41×10⁻⁶) N

I just have no idea how to do it with the y directed force since I only have a horizontal acceleration equation.
Thanks so much to anyone that helps, I appreciate it!
 
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Seriously, you're going to post something sideways so we have to get a kink in our neck looking at it?
 
phinds said:
Seriously, you're going to post something sideways so we have to get a kink in our neck looking at it?
Sorry I fixed it
 
Your problem is not well defined. What are h and L? What do they mean in the context of the problem? What does the net force (on what?) act on and how?'

You need to clearly state the entirety of the problem for Homework Helpers to provide meaningful help. Most will simply ignore your post otherwise; they don't have time to spare to play 20 questions as to what the actual problem statement is.
 
tj77 said:
Problem Statement: Assume the mass of the plane is 300,000kg h = 10 miles L = 25 miles and vxo = 100 mi/hr.
Determine the lift(or y-directed force not including the force of gravity) as a function of time.
Relevant Equations:

The question before asked to find the net force as a function of time which I got:

F = 4.83×10¹⁰ (1.65×10⁻⁸ t − 7.41×10⁻⁶) N


I think your final force is off by a factor of around 10. Did you accidentally add another zero or multiply it by something you shouldn't have?

I just have no idea how to do it with the y directed force since I only have a horizontal acceleration equation.
Thanks so much to anyone that helps, I appreciate it!

What makes you think y is in the horizontal direction? Look at the equations: y is proportional to h, and h typically denotes height.
 
collinsmark said:
I think your final force is off by a factor of around 10. Did you accidentally add another zero or multiply it by something you shouldn't have?
What makes you think y is in the horizontal direction? Look at the equations: y is proportional to h, and h typically denotes height.
I think that's what my teacher said, but I could've heard him wrong
 
tj77 said:
I think that's what my teacher said, but I could've heard him wrong
Ok, but you might want to re-think your horizontal vs. vertical. Anyway, double check your calculations or show your steps because I still think your final answer is off by about a factor of 10.
 
I did another post that might clear it up, thanks for your help I really appreciate it
 
tj77 said:
I did another post that might clear it up, thanks for your help I really appreciate it
Your last post didn't seem to clear things up. Let me help you out,

If h = 10 \ \mathrm{miles} = 16,093 \ \mathrm{m} and m = 300,000 \ \mathrm{kg}
then

mh \neq 4.83 \times 10^{10} \ \mathrm{m \ kg}
 
  • #10
collinsmark said:
Your last post didn't seem to clear things up. Let me help you out,

If h = 10 \ \mathrm{miles} = 16,093 \ \mathrm{m} and m = 300,000 \ \mathrm{kg}
then

mh \neq 4.83 \times 10^{10} \ \mathrm{m \ kg}
What equation would I use then?
 
  • #11
tj77 said:
What equation would I use then?
What is 16093 \times 300000?
 
  • #12
collinsmark said:
What is 16093 \times 300000?
Oh I see its 4.8 X 10^9. So that would be the answer?
 
  • #13
tj77 said:
Oh I see its 4.8 X 10^9. So that would be the answer?
I should be asking you that. :-p

But seriously, you should now be able to fix up your force equation in you initial post, and simplify it one final step.

All joking aside though, it would be beneficial for you to go through the problem and try to understand the meaning of each of the equations involved.

If you you would like more help, you can always show your steps complete with commentary on each step along with your impression of each of the original equations (i.e., what they mean, what they represent), and someone here will be happy to attempt help if there are any misconceptions.
 
  • #14
collinsmark said:
I should be asking you that. :-p

But seriously, you should now be able to fix up your force equation in you initial post, and simplify it one final step.

All joking aside though, it would be beneficial for you to go through the problem and try to understand the meaning of each of the equations involved.

If you you would like more help, you can always show your steps complete with commentary on each step along with your impression of each of the original equations (i.e., what they mean, what they represent), and someone here will be happy to attempt help if there are any misconceptions.
Ok thanks, will do
 
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  • #15
tj77 said:
Ok thanks, will do
collinsmark said:
I should be asking you that. :-p

But seriously, you should now be able to fix up your force equation in you initial post, and simplify it one final step.

All joking aside though, it would be beneficial for you to go through the problem and try to understand the meaning of each of the equations involved.

If you you would like more help, you can always show your steps complete with commentary on each step along with your impression of each of the original equations (i.e., what they mean, what they represent), and someone here will be happy to attempt help if there are any misconceptions.
I did it, if you would like to look :smile:
 
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  • #16
@tj77, please don't add "Homework HELP!" or similar to your threads. Everyone who is posting here in the Homework Fora is by default asking for help for homework problems, so titling your posts like the above, or adding it to your thread title doesn't provide us with any information.

BTW, yours is an unusual problem, with the weight of the aircraft given in kg, but the distances and speeds in miles and miles/hour. Also, that's a very slow speed for the plane. It would probably fall out of the sky at that speed. The landing speed for a Boeing 737 is somewhere between 150 to 200 mi/hr.
 
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