Force and tension questions?

Question

A paraglider is flying horizontally at a constant speed. Assume that only two forces act on it in the vertical direction, its weight and a vertical lift force exerted on its wings by the air. The lift force has a magnitude of 1900 N. For both questions, take the upward direction to be the +y direction.
(a) What is the magnitude and direction of the force that the paraglider exerts on the earth?


(b) If the lift force should suddenly decrease to 1000 N, what would be the vertical acceleration of the glider?

2)
A train consists of 50 cars, each of which has a mass of 6.5 103 kg. The train has an acceleration of 5.0 10-2 m/s2. Ignore friction and determine the tension in the coupling at the following places.
(a) between the 30th and 31st cars
N

(b) between the 49th and 50th cars





I appreciate the help, everyone. I basically need an online tutor haha.

Answer 1

The force on the earth is equal and opposite to that on the glider, 1900N + m*g. Since the glider is not accelerating in the vertical direction, the lift force must equal the weight forces, so m*g = 1900N The force on the earth is then 2*1900N, directed toward the glider.

If the lift force drops by 1000N, it is only 900N; the weight is still 1900N, so there is a net force downward of 1000N. The mass of the glider, from m*g = 1900N is 1900/g. The downward acceleration is then F/m = 1000*g/1900

Consider the 50th car. It is accelerating, along with the rest of the train, at a = 10^-2 m/sec^2. It's mass is m = 6.5*10^3 kg. Therefore it must be being pulled by a tension of T = m*a. The tension on the 49th car must pull two cars at that acceleration, or T = 2*m/a. So in general the tension between the nth and (n-1)th care is T(n) = (51-n)*m/a You can solve for n=31.

Answer 2

1a. The downward force is equal to the lift force (to keep
the glider in equilibrium). Hence,

Downward force on earth=1900N.

In this part of the problem, understand the concept of equilibrium.

1b. If the lift force is decreased to 1000N, there will be
an unbalanced force of 900N computed as (1900-1000). This will make the
glider drop at an acceleration a , which can be computed
using the formula, F=Ma, where F is the unbalanced
force in Newtons or 900N, M is the mass of the glider
in kg, and a the acceleration in m/s/s.

What's the mass of the glider, M? If the downward force equals 1900N as given in the first part of this problem, then M=1900/9.8 (from the formula Force=Mg).

Now substitute the given values in the formula F=Ma:

900=(1900/9.8)a
a=900/(1900/9.8)
=4.64m/s/s

In this part of the problem, understand the concept of unbalanced force in using the formula F=Ma. Also how to convert mass in kg to weight (which is a force) in Newtons.

2a. 20 cars are exerting a force on the rest of the 50 cars behind. The tension is equal to the unbalanced
force between these 2 groups of cars. What is the Force exerted by 20 cars. Use the formula F=Ma . F=20Ma.
What is the force exerted by the 30 cars? Use the same formula, F'=30ma. The tension in the coupling is equal
to the unbalanced force or F'-F=30Ma-20Ma=10Ma or
10*6.5*10^3*5*10^-2=3250Newtons.

2b. The tension is the unbalanced force between the 50th car and 49 cars.
Unbalanced force or F'-F=49Ma-Ma=48Ma or
48*6.5*10^3*5*10^-2=15,600N