Finding a plane's weight and horizontal acceleration at takeoff

[cbchapm2]

Homework Statement

On an airplane's takeoff, the combined action of the air around the engines and wings of an airplane exerts a 8240-N force on the plane, directed upward at an angle of 75.0° above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction.

What is the weight of the plane in N?

What is it's horizontal acceleration?

Homework Equations

Force in the x direction: Fx=Fcos(theta)
Force in the y direction: Fy=Fsin(theta)

To find the weight of the plane, I would first find the mass and then use the equation
W=m X g, g being -9.8 m/s^2.

To find acceleration in the x direction, I would use Ax=Fx/m

The Attempt at a Solution

I found Fx and Fy like so:
Fx=8240cos(75*)
=2132.67 N

Fy=8240sin(75*)
=7959.23 N

I'm stuck on what to do after that...if I try to use the original force given 8240 N in the F=mg equation to find the mass, I get 840.82 kg. If I were to try and plug that back into the W=mg equation, I would just get the same original force that I was already given.

I need the mass to find acceleration, and I know I would do so by using the Ax=Fx/m. I have the Fx, I just need to know how to find the mass.

Help?

Thanks!

 

[HallsofIvy]

The plane rises with constant vertical speed so the vertical force, which you say is 7959.23 N, must be exactly equal to the weight of the airplane, mass times g.

 

[cbchapm2]

So you're saying that the 7959.23 N is my weight of the plane? I guess that would make sense, since it's a constant vertical speed (y direction)...

 

[cbchapm2]

Got the right answer for both questions! Thanks so much!