Question

43. A rocket sled accelerates at a rate of 49.0m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

Answer 1

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

$F_x = m a_x$

where

$F_x$ is the horizontal component of the force

m is the mass of the passenger

$a_x$ is the horizontal component of the acceleration

Here we have

m = 75.0 kg

$a_x = 49.0 m/s^2$

Substituting,

$F_x=(75.0)(49.0)=3675 N$

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

$R=mg=(75.0)(9.8)=735 N$

where we used

$g=9.8 m/s^2$ as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

$F_y = 735 N$

and therefore the magnitude of the net force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N$

And the direction is given by

$\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}$