Calculating Runway Length and Tension for Transport Plane Takeoff

[Heat]

Homework Statement

A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is 700kg, and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to 2700 N . The tension in the towrope between the transport plane and the first glider is not to exceed 12000N.

If a speed of 40m/s is required for takeoff, what minimum length of runway is needed?

What is the tension in the towrope between the two gliders while they are accelerating for the takeoff?

Homework Equations

F=ma

The Attempt at a Solution

First I drew the problem out. On how I see it.

Now I will draw the forces on each object separately.

Transport Plane:

Normal and Weight equally the same.
Tension pulling from the left.
Force of 40m/s going to the right. (This is the boost it needs for the objects to move.)

Glider behind Transport Plane:

Normal and Weight equally the same.
Tension pulling from right larger.
Tension pulling from left smaller.

Glider behind Glider:

Normal and Weight equally the same.
Tension pulling from right larger.

 

[Doc Al]

Transport Plane:

Normal and Weight equally the same.
Tension pulling from the left.

Good.

Force of 40m/s going to the right. (This is the boost it needs for the objects to move.)

40 m/s is a speed, not a force! There's some unknown force pulling the Transport plane to the right. Luckily you won't need to know it.

Glider behind Transport Plane:

Normal and Weight equally the same.
Tension pulling from right larger.
Tension pulling from left smaller.

Good.

Glider behind Glider:

Normal and Weight equally the same.
Tension pulling from right larger.

Good.

Now assume the rope in the middle has the maximum tension allowed. Now figure out the acceleration. Hint: Consider the two gliders as a single system. What's the net force on those gliders? The total mass?

 

[Heat]

Total Mass would be 1400kg.
Net force would be 2(tension pulling from Transport Plane).

If this is correct, I will go on to the next step.
The basics I just learned from you. :)

The part I don't understand is now that I have an "single" object of 1400kg, the weight and normal force cancel out and the only horizontal force left is unknown. :O

The transport plane has also a weight and normal force that cancel out, the boost it's getting is larger than the tension from the left. But for this one I don't got values. :(

I went a little ahead and know that once I have the acceleration, I will be able to plug it into this equation. v^2 = vi^2 + 2a(x-xi)

 

[Doc Al]

Total Mass would be 1400kg.
Net force would be 2(tension pulling from Transport Plane).

The net force on the gliders is that single tension from the rope attached to the Transport Plane. And that tension is given! (The rope between the gliders pulls in both directions, so its tension cancels out when looking at the system of both gliders together.) (Edit: Oops--my bad! We forgot the resistance acting on each glider, which is also given.)

If this is correct, I will go on to the next step.
The basics I just learned from you. :)

The part I don't understand is now that I have an "single" object of 1400kg, the weight and normal force cancel out and the only horizontal force left is unknown. :O

But the horizontal force is not unknown.

The transport plane has also a weight and normal force that cancel out, the boost it's getting is larger than the tension from the left. But for this one I don't got values. :(

This is true. Luckily, you know that all three are attached and must accelerate together. Thus if you can figure out the acceleration just by analyzing the gliders, that acceleration applies to everything.

I went a little ahead and know that once I have the acceleration, I will be able to plug it into this equation. v^2 = vi^2 + 2a(x-xi)

Excellent. That's the one you need.

 

[Heat]

The tension would be positive, gravity is pulling down so it would be -mg and acceleration is +a.

T-mg=a

1200-1400(9.8)= -12520

would the acceleration be correct, but then why is it negative. :S

let me attempt it again. :)

Update, I know why, Gravity is a vertical component here, we just need to calculate horizontal.

So this should be F=ma

1200 = 1400(a)
a= .857

and if this is true, then the distance should be 933.49, converting to 2 sig figs it would be 9.3x10^2

 

[Doc Al]

let me attempt it again. :)

Update, I know why, Gravity is a vertical component here, we just need to calculate horizontal.

So this should be F=ma

1200 = 1400(a)
a= .857

and if this is true, then the distance should be 933.49, converting to 2 sig figs it would be 9.3x10^2

Right approach, but we need the correct net force on the gliders. In my last post, I forgot about the resistive force that acts on the gliders, which is also given.

In light of that, what's the real net force on the gliders? (Also: The towrope tension is 12,000 N not 1200 N.)

 

[Heat]

The resistive force is 2700. (This is where it confuses me, does this include the transport plane, or just the gliders.)

12000-2700 = 9300
9300 = 1400a
a= 6.64 m/s^2

 

[Doc Al]

The resistive force is 2700. (This is where it confuses me, does this include the transport plane, or just the gliders.)

Reread the problem statement:

The mass of each glider is 700kg, and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to 2700 N.

 

[Heat]

got it so the friction should be 2700. friction should double.

12000-2(2700) = 6600
6600 = 1400a
a = 4.71

 

[Doc Al]

Looks good.

 

[Heat]

Ok, I managed to get the part a, with the kinematic equation. :D

Now for the tension in the towrope between the two gliders while they are accelerating for the takeoff.

We separate the blocks/objects again,
we know their mass of 700kg each, the acceleration, and we find tension.

T=ma
T = 700(4.71)?

 

[Doc Al]

When applying Newton's 2nd law, you must consider all the forces acting on the glider.

 

[Heat]

I thought for the tension (x component) was important only. Gravity is vertical here, and the x component of gravity is 0.

Wait, I just remembered the resistance of 2700.

I will see if I could work this out. :)

 

[Heat]

1500N is the the net force.

T - 1500 = 700(6.64)
T = 3148

 

[Doc Al]

1500N is the the net force.

T - 1500 = 700(6.64)
T = 3148

First off, the net force is the sum of all forces acting on the glider. There are two: The tension (T) to the right, and the resistive force (which is given) to the left. Add them and set that sum equal to ma.

How did the acceleration change from what you had calculated earlier?

 

[Heat]

lol, sorry the acceleration was my mistake. I have scrolled up too much, that I wrote down the wrong acceleration.

anyways, I thought net force was the final force. The weight and normal force cancel out, why do they not add up (is it because of opposite direction). Same goes to the object, it's getting pulled to the right, but resistance to the left, hence why I thought it was the difference.

So Fnet = 14700
a = 4.71
m = 700

T = 14700+ (700*4.71)
T = 17997

 

[Doc Al]

anyways, I thought net force was the final force. The weight and normal force cancel out, why do they not add up (is it because of opposite direction).

The weight (-mg) and normal force (+mg) do add up--to zero! So forget them.

Same goes to the object, it's getting pulled to the right, but resistance to the left, hence why I thought it was the difference.

Yes, the net force is the difference (or sum): +T - F(resistive) = T - 2700. Set that equal to ma.

So Fnet = 14700
a = 4.71
m = 700

T = 14700+ (700*4.71)
T = 17997

Where does the 14700 come from? Do over!

 

[Heat]

T - 2700 = 700(4.71)
T = 2700 + 3297
T = 5997

I just understand that it was +T and -F. This should be right now. If it is, I would like to thank you Doc Al.

 

[Doc Al]

Looks good to me. (But don't forget units when submitting your answer.) Why don't you mark this one as "solved"?

 

[Heat]

how do I do that?

 

[Doc Al]

how do I do that?

Go to "Thread Tools" and you'll find an option for doing that.

 

[Heat]

I see the following
Thread Tools:
Show Printable Version
Email this Page
Subscription/Unsubscribe from this Thread
Add a Poll to this Thread

 

[Doc Al]

Hmmm... you don't see an option labeled: Mark this thread as solved?

 

[Heat]

nope,

I am using Firefox btw.