Explanation:
It is given that,
Mass of the passenger, m = 75 kg
Acceleration of the rocket, [tex]a=49\ m/s^2[/tex]
(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :
F = m a
[tex]F=75\ kg\times 49\ m/s^2[/tex]
F = 3675 N
Ratio, [tex]R=\dfrac{F}{W}[/tex]
[tex]R=\dfrac{3675}{75\times 9.8}=5[/tex]
So, the ratio between the horizontal force and the weight is 5 : 1.
(b) The magnitude of total force the seat exerts against his body is F' i.e.
[tex]F'=\sqrt{F^2+W^2}[/tex]
[tex]F'=\sqrt{(3675)^2+(75\times 9.8)^2}[/tex]
F' = 3747.7 N
The direction of force is calculated as :
[tex]\theta=tan^{-1}(\dfrac{W}{F})[/tex]
[tex]\theta=tan^{-1}(\dfrac{1}{5})[/tex]
[tex]\theta=11.3^{\circ}[/tex]
Hence, this is the required solution.
