# Horizontal acceleration of a plane with constant vertical velocity

1. Sep 19, 2008

### WHarmon

1. The problem statement, all variables and given/known data
On takeoff, the combined action of the air around the engines and wings of an airplane exerts a 7049N force on the plane directed upward at an angle of 58.7 above the horizontal. The plane rises with a constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. The acceleration of gravity is 9.8m/s2.
What is the weight of the plane?
What is the horizontal acceleration of the plane?

2. Relevant equations
Net force = ma

3. The attempt at a solution
I determined the weight of the plane to be 6023N using newtons second law and because of the lack of the vertical acceleration it was rather simple. Now I cannot find the horizontal acceleration. I have the thrust pushing the plane forward (7049N) and the weight of the plane (6023N), but I am still stuck with two variables, my restrictive force and the acceleration. Is there something that I am missing when it comes to determining the restrictive force?

2. Sep 20, 2008

### Kurdt

Staff Emeritus
Well it seems you've calculated the vertical component of the force just fine, what is wrong with calculating the horizontal component? After that you can use Newton's second law to find the acceleration.

3. Sep 20, 2008

### WHarmon

Calculating the vertical forces wasn't a problem because the acceleration in that direction was understood to be 0 thus making it an equilibrium problem. I am stuck on the horizontal forces because I have 2 unknowns and I cannot find a way to figure either of them out.

I know that my [net force = the thrust of my plane - restrictive force - weight = mass*acceleration in the x direction]

condensed that would be something like:
T - R - W = ma

And I just don't know where to go from here.

4. Sep 20, 2008

### Kurdt

Staff Emeritus
5. Sep 20, 2008

### WHarmon

Yes I broke the thrust force down into x and y components. The y component was the equivalent of the normal force and thus equal to the weight. This is how I answered the first half of the problem. I also have the x component of the force, but not the force that opposes this or the acceleration in this direction. Because the object is not at equilibrium in the x direction I cannot assume that the net force is equal to 0.

6. Sep 20, 2008

### Kurdt

Staff Emeritus
There is no force that opposes it. You have not been asked to consider drag or anything like that, therefore to find the acceleration you use Newton's second law on the x-component of the force.

7. Sep 20, 2008

### WHarmon

So then what I am left with is

T - W = ma?

8. Sep 20, 2008

### Kurdt

Staff Emeritus
You just need the thrust in the x direction is equal to ma.

9. Sep 20, 2008

### WHarmon

Ok, that gave me the correct answer, but i would like to be sure that I have a firm grasp on this before i quit monitoring this thread.

I am still confused why you don't include the weight in your net force. Is it because your thrust force is at an angle and you are only using the horizontal component? Does this not call for the use of any vertical force?

Had my problem been dealing with a car being pushed and accelerated on a horizontal plane would I incorporate the weight and frictional force in my answer?

10. Sep 20, 2008

### Kurdt

Staff Emeritus
Yes the vertical and horizontal components should be treated separately.

If you were asked to consider friction in the car question then yes the weight of the car would be proportional to the amount of friction there was.