Your spacecraft has two engines, on the left and the right sides, but the right

Question

Your spacecraft has two engines, on the left and the right sides, but the right engine

was destroyed by a proton torpedo. Using the remaining engine by itself would create

a large torque that would cause your ship to rotate and thus y in circles, keeping you

from ever getting back to Earth. So in addition to the engine, you must use the three

small control jets on your ship to keep it ying straight. Using the center of mass of

the ship as the origin, the working engine is positioned at (??2;??1; 0), and produces a

thrust of 215 kN in the ~j direction. The control jets are arranged as follows:

Control jet A, located at (1; 10; 1), produces thrust in the ??~i

direction.

Control jet B, located at (8; 2;??3), produces thrust in the ~k direction.

Control jet C, located at (6;??1; 2), produces thrust in the ~j direction.

How much thrust must each control jet produce in order to prevent the ship from

spinning? (You are encouraged to use a calculator, or even a computer, to help solve

this.)

With the forces you found, in what direction will the spaceship accelerate?

Assuming the ship has a mass of 732 kg, what will be the magnitude of the resulting

acceleration?

Explanation / Answer

Let thrust produce by Control jet A is Fa;

by Control jet B is Fb;

by Control jet C is Fc;

To prevent ship from spinning, torque about centre of mass should be zero.

hence balancing torque by every control jet and engine about centre of mass,we have

Torque by engine

(-2i-j)X(215j)=-430k

Torque by Control jet A

(i+10j+k)X(-Fa i)=-Fa j + 10Fa k

Torque by Control jet B

(8i+2j-3k)X(Fb k)= 2Fb i - 8Fb j

Torque by Control jet C

(6i-j+2k)X(Fc j)=-2Fc i + 6Fc k

hence sum of all these torque should be zero.

(2Fb-2Fc)i +(-8Fb-Fa)j + (10Fa+6Fc-430) =0.

Solving this we have,

Fa=48.486 kN

Fb=-5.81 kN

Fc= -5.81kN.

Using these values

net force on ship(F)= 48.486 i +(215-5.81) j - 5.81 k

=48.486 i +(209.19) j - 5.81 k kN

|F|=214.81 KN.

So it will go in direction along unit vector=F/|F|=(48.486 i +(209.19) j - 5.81 k)/214.81

accelaration ship=|F|/m=214.81*1000/732 m/s^2 = 293.46 m/s^2.

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