Q11PE - College Physics (Urone)

Question

The rocket sled shown in Figure 4.33 accelerates at a rate of 49.0 m/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 by using a ratio. 

(b) Calculate the direction and magnitude of the total force the seat exerts against his body.

Step-by-Step Solution

Answer

(a) The horizontal component of the force exerted is 3675 N.

(b) The magnitude of the net force exerted by the seat is 3747.8 N, and the direction 

       of the net force with the horizontal is $11.{31}^{\circ }$.

Step 1: Determine the horizontal component of the force.

Apply Newton’s second law of motion.

F_h = ma ............(1)

Here, F_h is the horizontal component of force, m is the mass of the passenger, and a is the acceleration.

Step 2: Given Data

• Acceleration =  49.0 m/s²
• Mass of the austronaut = 75.0 kg.

Step 3: Calculating the horizontal component of force

(a)

Substitute 75 kg for m and 49.0 m/s² for a in the equation (i), and we get,

$\begin{array}{l}{F}_h=75\mathrm{kg}\times 49 m/s^2\\ =3675 \mathrm{kg} m/s^2\\ =3675 N\end{array}$

Hence, the horizontal component of the force is 3675 N.

Step 4: Determine the ratio of the horizontal component with the weight of the passenger

Calculate the weight of the passenger as:

 

W = mg

Here, W is the weight of the passenger, and g is the acceleration due to gravity.

 

Substitute 75 kg for m and 9.8 m/s² for g in the above expression, and we get,

$\begin{array}{ll}W=75 \mathrm{kg}\times 9.8m/s^2& \\ = 735 kg. m/s^2& \\ =735N& \end{array}$

Determine the ratio of the horizontal component of a force and the weight of the passenger as,

$$\begin{gathered} \frac{F_h}{W} = \frac{3675 N}{735 N} \\ = 5 \end{gathered}$$

Hence, the horizontal component of force is five times greater than the weight of the passenger.

Step 5: Determine the magnitude and direction of the force exerted by the seat.

(b) 

Draw the free-body diagram of the forces acting as shown below.

Here, N is the normal reaction, $\theta$ is the direction of the net force with the horizontal, and F_net is the net force.

The normal force is equal to the weight.

 

Calculate the net force as:

$F_{net}=\sqrt{\left(F_h^2+N^2\right)}$

Substitute 3675N for F_h and 735N for N in the above expression, and we get,

$$\begin{gathered} F_{net} = \sqrt{\left({3675}^2+{735}^2\right)N} \\ {F}_{net} = \sqrt{\left(13505625 + 540225\right)N} \\ {F}_{net} = \sqrt{\left(14045850\right)N} \\ {F}_{net} = 3747.8 N \end{gathered}$$

Hence, the magnitude of the net force exerted by the seat is 3747.8 N.

 Determine the direction of the net force as,

$\tan \theta =\frac{N}{F_h}$

 Substitute 3675N for F_h and 735N for N in the above expression, and we get,

$\begin{array}{l}{\tan }^{\theta }=\frac{735N}{3775N}\\ \theta ={\tan }^{-1}\left(02\right)\\ \theta =11. {31}^{\circ }\end{array}$

Hence, the direction of the net force with the horizontal is $11.{31}^{\circ }$.