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Force exerted at given angle, constant velocty, find weight

by AR8742
Tags: angle, constant, exerted, force, velocty, weight
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AR8742
#1
Sep13-08, 09:49 PM
P: 3
1. The problem statement, all variables and given/known data
On takeoff, the action of the air around the engine and wings of an airplane exerts a 7546 N force on the plane, directed upward at an angle of 70.5 degrees above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. Acceleration due to gravity is 9.8m/s^2. What is the weight of the plane (in N)?


2. Relevant equations
I think I should use F=ma. (?)


3. The attempt at a solution
My first problem is I'm not quite sure if I have the free body diagram right. I have the plane's weight in the downward direction, normal force in the upward direction, and the (7546 N) force in between the normal force and the horizontal axis at 70.5 degrees. It's hard to know if I'm on the right track if that isn't right. From there I simply tried m=F/a. m = 7546N / 9.8m/s^2 = 770 kg? That would be too easy though and I feel like a calculation for acceleration is necessary? My previous problems involved finding acceleration with friction forces so I am a bit thrown off by this for some reason. Do I have to find the sum of the forces? Just looking if someone can point me in the right direction either with the free body diagram or an equation. Thanks!
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alphysicist
#2
Sep14-08, 12:15 AM
HW Helper
P: 2,249
Quote Quote by AR8742 View Post
1. The problem statement, all variables and given/known data
On takeoff, the action of the air around the engine and wings of an airplane exerts a 7546 N force on the plane, directed upward at an angle of 70.5 degrees above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. Acceleration due to gravity is 9.8m/s^2. What is the weight of the plane (in N)?


2. Relevant equations
I think I should use F=ma. (?)


3. The attempt at a solution
My first problem is I'm not quite sure if I have the free body diagram right. I have the plane's weight in the downward direction, normal force in the upward direction, and the (7546 N) force in between the normal force and the horizontal axis at 70.5 degrees.
Since they say the plane is rising, it has already left the ground, so one of these forces should is not present.

It's hard to know if I'm on the right track if that isn't right. From there I simply tried m=F/a. m = 7546N / 9.8m/s^2 = 770 kg?
I think you need to be a bit more careful with Newton's law here. If you pick some specific direction (and call it the x direction), then it is:

[tex]
\sum F_x = m a_x
[/tex]

that is, add up all of the x components of the forces acting on the object, and then that's equal to mass times the x component of the acceleration.

So what is the best direction to apply Newton's law here, based on what they are asking? What do you get?
AR8742
#3
Sep14-08, 01:57 AM
P: 3
Ok, that's what I was thinking...except I just wasn't sure I had all of the forces right in my free body diagram. So there would be no normal force then (just weight and the exerted force)? If I make the +x direction on the horizontal axis my sum would be Fcos70.5 = (ma) in x direction and my sum in the +y direction would be Fsin70.5-w =(ma)in y direction. But at this point I am still confused because I don't know the plane's weight or acceleration. I'm sure this problem isn't that difficult but for some reason nothing is clicking for me about where to go from here.

alphysicist
#4
Sep14-08, 02:13 AM
HW Helper
P: 2,249
Force exerted at given angle, constant velocty, find weight

Quote Quote by AR8742 View Post
Ok, that's what I was thinking...except I just wasn't sure I had all of the forces right in my free body diagram. So there would be no normal force then (just weight and the exerted force)? If I make the +x direction on the horizontal axis my sum would be Fcos70.5 = (ma) in x direction and my sum in the +y direction would be Fsin70.5-w =(ma)in y direction. But at this point I am still confused because I don't know the plane's weight or acceleration. I'm sure this problem isn't that difficult but for some reason nothing is clicking for me about where to go from here.
You're trying to find the planes weight, so the y-equation is the one to use. What is the acceleration in the y direction?
AR8742
#5
Sep14-08, 04:02 PM
P: 3
Is it 9.8m/s^2?
alphysicist
#6
Sep14-08, 05:19 PM
HW Helper
P: 2,249
Quote Quote by AR8742 View Post
Is it 9.8m/s^2?
No, but you can determine the vertical acceleration from the fact that it says the plane rises with constant velocity in the vertical direction. What would that be?


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