Question

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 700 kg, and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to 4300 N. The tension in the towrope between the transport plane and the first glider is not to exceed 12000 N.

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

Answer 1

Using Newton's second law and kinematic equations, the minimum length of the runway needed for the gliders to take off at a speed of 40 m/s, without exceeding the maximum tow rope tension, is calculated to be 3.64 meters.

Step-by-step explanation:

To find the minimum length of the runway needed for the gliders to take off, we need to calculate the acceleration that the tow rope can provide without breaking. We also need to calculate the distance required to reach the takeoff speed under this constant acceleration.

First, we use Newton's second law, F = ma, where F is the net force, m is the mass, and a is the acceleration. For one glider, the maximum tension the tow rope can provide is 12,000 N, and the resistance is 4,300 N. Therefore, the maximum net force on the glider is 12,000 N - 4,300 N = 7,700 N.

Now we can calculate the acceleration: a = F/m, which gives us a = 7,700 N / 700 kg = 11 m/s².

To find the distance, we use the kinematic equation v² = u² + 2ad, where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration, and d is the distance. With a takeoff speed of 40 m/s required, we get 40 m/s² = 0 + 2 * 11 m/s² * d, which simplifies to d = 40 m/s² / (2 * 11 m/s²) = 80 m/s² / 22 m/s² = 3.64 m.

Therefore, the minimum length of the runway needed for the gliders to take off is 3.64 meters.

Answer 2

Length = 155.6 m

Step-by-step explanation:

given data

mass = 700 kg

total resistance = 4300 N

tension = 12000 N

speed = 40 m/s

solution

we get here equation for glider in the back and front

for the glider in the back

T - 2400 = 700 a ...........1

for the glider in front

12000 - T - 2400 = 700a .............2

now we add both these equations

12000 - 4800 = 1400 a .............3

a = 5.14 m/s²

and

now we use equation of motion

v² - u² = 2 a S .............4

40² = 2 × 5.14 × S

so

Length = 155.6 m