Physics help!

letsbefree

Canada123 Posts

August 29 2008 15:23 GMT

I'm doing this monstrous paper filled with hardcore physics problems and I'm having a bit of difficulty, hope you guys can help me out:

Two questions:
1) An airplane wishes to fly from Toronto to Ottawa. Toronto is a distance of 320 km S40W from Ottawa. If the plane can fly with airspeed of 600 km/hr and there has been a wind speed of 75 km/hr coming from the southwest measured by the weather network. What is the direction the plane should point from T.O to arrive in Ottawa, and how long will it take?

2) An airplane is flying in a horizontal circle at a speed of 480 km/h. If the wings are tilted 40 degrees from the horizontal, what is the radius of the circle in which the plane is flying? Assume that the force opposing gravity to keep the plane in the air is horizontal lift which exactly balances the force of gravity when the plane is flying straight.

Thanks

Chill

Calgary25980 Posts

August 29 2008 15:34 GMT

You know, I'd like to see you try the problem or ask specific questions, rather than have me do all this work for you. I'm sure there are similar solutions online that you can copy the process from.

MasterOfChaos

Germany2896 Posts

August 29 2008 15:43 GMT

1) Approx: Earth is flat
x Vector start-goal
x2=x-vw*t Vector start-effective goal
vw Vector windspeed
vp scalar planespeed
x2^2=vp^2
solve for t
then calc x2
then use arctan to get angle from x2 vector

micronesia

United States24676 Posts

August 29 2008 16:36 GMT

I agree with Chill.

Also, state your exigency! What level is this for?

skyglow1

New Zealand3962 Posts

August 29 2008 16:47 GMT

For 1 you could do it by vector components and trigonometry:

Let A be the resultant speed of the plane and x be the angle clockwise from south which the plane has to fly

For the 75kmh wind:
x and y components are both 75cos45

For the plane:
x component is 600sinx
y component is 600cosx

For the resultant:
x component is Asin40 (we want it to fly in direction S40W)
y component is Acos40

Wind + plane = resultant
For x components: 75cos45-600sinx = -Asin40 (taking east and north as positive, and putting in the negative signs knowing that those components will be pointing west and south)
For y components: 75sin45-600cosx = -Acos40

Rearranging both equations for A:
A = (75cos45-600sinx)/-sin40
A = (75sin45-600cosx)/-cos40
So (75cos45-600sinx)/-sin40 = (75sin45-600cosx)/-cos40
Multiplying out: -75cos45cos40+600cos45sinx = -75sin45sin40+600sin40cosx
Tidying up: (75sin45sin40-75cos45cos40)/600 = sin40cosx-cos40sinx

Using the identity acosx-bsinx = R(cos+alpha) where R is sqrt(a^2+b^2) and alpha = arctan(b/a):
R = 1 and alpha = 50
So sin40cosx-cos40sinx = cos(x+50) and:
(75sin45sin40-75cos45cos40)/600 = cos(x+50)
Solving for x gives 40.62 degrees
So the plane needs to fly at a direction that is 40.62 degrees clockwise from south.

To find A, the substitue x into the x and y components of the resultant:
x component is -337.6 and y component is -402.4 so using pythogoras A = 525.2km/h
Then the time it'll take to reach there is 0.609 hours or 36.6 minutes.

Messy way, but it gets the job done

(assuming I didn't screw up anywhere)

skyglow1

New Zealand3962 Posts

August 29 2008 17:14 GMT

For 2:
Let T be the force provided by the wing. T is tilted 40 degrees from straight up, so:
x component is Tsin40
y component is Tcos40

Let m be the mass of the plane:
Then Tcos40 = mg because the vertical forces are balanced

From a right angle triangle containing 40 degrees as an angle and with Tcos40, T sin40 and T as the sides we can write:
tan40 = Tsin40/Tcos40
So Tsin40 = tan40 x Tcos40
But Tcos40 = mg so substituting gives:
Tsin40 = tan40 x mg

F = mv^2/r for centripetal force
F is the x component which is Tsin40 which is tan40 x mg so:
tan40 x mg = mv^2/r
the m's cancel out so tan40 x 9.81 = (480/3.6)^2/r
Solving for r gives 2.16km

letsbefree

Canada123 Posts

August 29 2008 17:19 GMT

On August 30 2008 00:34 Chill wrote:
You know, I'd like to see you try the problem or ask specific questions, rather than have me do all this work for you. I'm sure there are similar solutions online that you can copy the process from.

i didn't intend to let you guys do all the work. =( I'll be careful from now on. Sorry.

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