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Runway [ Tension ] 
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#1
Oct607, 08:54 PM

P: 273

1. The problem statement, all variables and given/known data
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? 2. Relevant equations F=ma 3. 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. 


#2
Oct607, 09:04 PM

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P: 41,477

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? 


#3
Oct607, 09:09 PM

P: 273

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(xxi) 


#4
Oct707, 06:46 AM

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P: 41,477

Runway [ Tension ]



#5
Oct707, 08:50 AM

P: 273

The tension would be positive, gravity is pulling down so it would be mg and acceleration is +a.
Tmg=a 12001400(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 


#6
Oct707, 10:21 AM

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P: 41,477

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


#7
Oct707, 10:25 AM

P: 273

The resistive force is 2700. (This is where it confuses me, does this include the transport plane, or just the gliders.)
120002700 = 9300 9300 = 1400a a= 6.64 m/s^2 


#9
Oct707, 10:32 AM

P: 273

got it so the friction should be 2700. friction should double.
120002(2700) = 6600 6600 = 1400a a = 4.71 


#11
Oct707, 10:47 AM

P: 273

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)? 


#12
Oct707, 10:49 AM

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P: 41,477

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



#13
Oct707, 10:52 AM

P: 273

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. :) 


#14
Oct707, 12:06 PM

P: 273

1500N is the the net force.
T  1500 = 700(6.64) T = 3148 


#15
Oct707, 12:11 PM

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P: 41,477

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


#16
Oct707, 12:17 PM

P: 273

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 


#17
Oct707, 12:45 PM

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P: 41,477




#18
Oct707, 12:48 PM

P: 273

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. 


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