Vectors and Forces

  • Thread starter CathyLou
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Would anyone please be able to help me with the following question. I'd really appreciate it.

6. An aircraft of mass 11000kg, which moves at a constant velocity v and constant altitude, is powered by propellors and experiences a drag force.

(a) Draw a labelled free body diagram showing the four main forces acting on the aircraft. (I don't know how to draw the diagram, but I think that the 4 forces are gravity, air resistance, friction and upthrust. Is this correct?)

(b) The thrust from the propellors is 225kN and the drag force is given by 10v^2. Calculate the aircraft’s level flight speed.


Thank you.

Cathy
 

Answers and Replies

  • #2
Hootenanny
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CathyLou said:
Would anyone please be able to help me with the following question. I'd really appreciate it.

6. An aircraft of mass 11000kg, which moves at a constant velocity v and constant altitude, is powered by propellors and experiences a drag force.

(a) Draw a labelled free body diagram showing the four main forces acting on the aircraft. (I don't know how to draw the diagram, but I think that the 4 forces are gravity, air resistance, friction and upthrust. Is this correct?)
There are indeed four forces acting but one of your forces is wrong, you need to replace that force with another. Can you figure out which one it is? Remember constant velocity implies no net force.
HINT: The answer is in the next question :wink:
CathyLou said:
(b) The thrust from the propellors is 225kN and the drag force is given by 10v^2. Calculate the aircraft’s level flight speed.
Have you any thoughts on this one? Exactly the same principles apply here as above.
 
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  • #3
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Thanks for replying.

Is it friction that is wrong? If yes, does that mean that there is a lift force that opposes the force of gravity instead. Also, I don't really get what a free body diagram is. Could you please explain?

As for part b, if thrust is equal to 225kN the drag must be 225kN.

So, 10v^2 = 225kN

and 10v^2 = 225000N

so v^2 = 22500N

and v = 150m/s

Is that correct?
 
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  • #4
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CathyLou said:
Is it friction that is wrong? If yes, does that mean that there is a lift force that opposes the force of gravity instead. Also, I don't really get what a free body diagram is. Could you please explain?
I've figured out how to draw a free body digaram now. :smile:
 
  • #5
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CathyLou said:
Is it friction that is wrong? If yes, does that mean that there is a lift force that opposes the force of gravity instead. Also, I don't really get what a free body diagram is. Could you please explain?
Yes, friction is "wrong," i.e., air resistance is a form of friction. What you labelled as upthrust is known as lift, and the missing force that Hootenanny was talking about is the (forward) thrust.

As for part b, if thrust is equal to 225kN the drag must be 225kN.

So, 10v^2 = 225kN

and 10v^2 = 225000N

so v^2 = 22500N

and v = 150m/s

Is that correct?
That's right!
 
  • #6
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CathyLou said:
I've figured out how to draw a free body digaram now. :smile:
Good for you. :)
 
  • #7
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neutrino said:
Good for you. :)
:smile:

I just posted that so that no-one spent time showing me as there was no longer any need (thanks to Wikipedia!).

Cathy
 
  • #8
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neutrino said:
Yes, friction is "wrong," i.e., air resistance is a form of friction. What you labelled as upthrust is known as lift, and the missing force that Hootenanny was talking about is the (forward) thrust.


That's right!
Thanks for your help!
 
  • #9
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I'm also kind of stuck on the following question.

Could anyone please offer some hints?

2. A ship is pulled at a constant speed, v, of 2.5 m/s by two tugs, A and B. Each tug is connected to the ship by a cable so that the angle each of the cables makes with the direction of travel is 41 degrees. The ship experiences a drag force given by:

Drag Force (N) = 8000 x velocity^2

(a) Calculate the tension in each cable while travelling at this constant speed. (I got an answer of 66200 N for each tug but I'm not sure if that's correct).

(b) As the tugs attempt to increase the speed of the ship from 2.5m/s, tug A breaks down, with its cable to the ship becoming slack.

<i> Calculate the speed to which the ship initially decelerates, assuming the tension in the other cable remains constant. (Go backwards and work out v through friction force).

<ii> The ship also veers off-course. Explain why this happens.


I've already drawn a diagram.

Thank you.

Cathy
 
  • #10
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Right, I've figured out part a to be T = 33100 N.

Does anyone have any idea how to work out part b?

Thank you.

Cathy
 
  • #11
radou
Homework Helper
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CathyLou said:
...
(b) As the tugs attempt to increase the speed of the ship from 2.5m/s, tug A breaks down, with its cable to the ship becoming slack.

<i> Calculate the speed to which the ship initially decelerates, assuming the tension in the other cable remains constant. (Go backwards and work out v through friction force).

<ii> The ship also veers off-course. Explain why this happens.[/COLOR]
Obviously the speed isn't constant anymore. What does this imply?
 
  • #12
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radou said:
Obviously the speed isn't constant anymore. What does this imply?
That the forces aren't balanced???
 
  • #13
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I've figured out b now but I still can't work out c.

Any suggestions as to how to do it?

Thank you.

Cathy
 
  • #14
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I've now got an answer to part c.

Thanks for everyone's help.

I really appreciate it.

Cathy :smile:
 

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