# Forces involved in a climbing jet aircraft

I have a problem with which I need help. The question is:

A jet aircraft is climbing at an angle of 45° above the horizontal and is accelerating at 4.5m/s2. What is the total force that the cockpit seat exerts on the 75kg pilot?

It says that the y-component of the force is:

$$F\sin(a)–mg\cos(45^\circ) = 0.$$

I don't understand how this component is working at all. I was just wondering if anybody could help me understand why this is the y-component of the force exerted by the seat.

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Ever heard of a Free Body Diagram? Look it up. –  ja72 Oct 31 '13 at 14:31

You just consider the forces in the direction orthogonal to the trajectory (I guess this is the y direction). According to the conditions of the problem, there is no acceleration along this direction, so the sum of the projections of all forces onto this direction is zero. The forces are gravity and the force exerted by the seat.

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First, find the forces that you need.

The pilot is accelerating at $4.5 {m}/{s^2}$ at 45 degrees above the horizontal. The net horizontal and vertical (upward) forces that must be acting on the pilot, to produce this acceleration, are both $4.5\times 75 \times sin(45)$ (since $\sin(45) =\cos(45)$)

Next, look at the forces you have.

Gravity is pulling down on the pilot with $9.8\times 75 \$ newtons. The seat must then be exerting the exact force that, added to the force of gravity, gives the required result.

So, the force from the seat must be $(4.5\times 75 \times sin(45) + 9.8\times 75 \ )$ newtons vertically upward, and $(4.5\times 75 \times sin(45))$ newtons horizontally.

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You have to add the accelerations as vectors. You have 4.5 m/s2 at 45° up and left, and 9.8 m/s2 up. Add the two, then take the magnitude of the resulting vector.

$(-3.18, 3.18) + (0, 9.8) = (-3.18, 12.98)$

The magnitude of that vector is 13.4, so the total apparent acceleration experienced by the pilot is 13.4 m/s2. The force exerted on the pilot is therefore 13.4 m/s2 x 75 kg = 1.0 kN.

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