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=√(F_x^2+F_y^2)=√(3675^2+735^2)=3748 N`

And the direction is given by

`\theta = tan^(-1)((F_y)/(F_x))=tan^(-1)((735)/(3675))=11.3^(\circ)`