Chapter 4, Problem 157P

(a)

To determine

The direction of force exerted by the air on the airplane.

Answer to Problem 157P

The force is directed at $\text{5}\text{.0°}\text{from}\text{the}\text{vertical}$.

Explanation of Solution

Mass of airplane is $2800\text{kg}$, velocity of altitude gain is $2.3\text{m}/\text{s}$, and the acceleration of horizontal component of velocity is $0.86\text{m}/{\text{s}}^{\text{2}}$.

Write the equation to find the angle made by force with the vertical.

$\theta ={\tan }^{-1}\left(\frac{F_x}{F_y}\right)$

Here, the angle made by the force with vertical is $\theta$, x-component of force is $F_x$, and the y-component of fore is $F_y$.

Write the expression to find $F_x$.

$F_x=m{a}_x$

Here, the mass of airplane is $m$, and the x-component of acceleration is $a_x$.

Write the expression to find $F_y$.

$F_y=mg$

Here, the gravitational acceleration is $g$.

Rewrite the equation for $\theta$ by substituting the relations for $F_x$ and $F_y$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{m{a}_x}{mg}\right)\\ ={\tan }^{-1}\left(\frac{a_x}{g}\right)\end{array}$

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.86\text{m}/{\text{s}}^{\text{2}}$ for $a_x$ in the above equation to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0.86\text{m}/{\text{s}}^{\text{2}}}{9.8\text{m}/{\text{s}}^{\text{2}}}\right)\\ ={\tan }^{-1}\left(0.087\right)\\ =5.0°\end{array}$

Therefore, the force is directed at $\text{5}\text{.0°}\text{from}\text{the}\text{vertical}$.

(b)

To determine

The horizontal and vertical components of acceleration of airplane.

Answer to Problem 157P

X-component of acceleration is $0.51\text{m}/{\text{s}}^{\text{2}}$.

Y-component of acceleration is $0.02\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Mass of airplane is $2800\text{kg}$, velocity of altitude gain is $2.3\text{m}/\text{s}$, acceleration of horizontal component of velocity is $0.86\text{m}/{\text{s}}^{\text{2}}$, and the force exerted by air has direction $2.0°$ closer to vertical than in part (a).

Write the equation for the magnitude of net acceleration of plane.

$a=\sqrt{a_x^2+g^2}$

Here, the magnitude of net acceleration of plane is $a$.

Write the equation for the x-component of acceleration of plane.

$a_x=a\sin \theta$

Here, the magnitude of acceleration is $a$.

Rewrite the equation for $a_x$ by substituting relation for $a$.

$a_x=\left(\sqrt{a_x^2+g^2}\right)\sin \theta$ (I)

Write the equation for y-component of acceleration of plane.

$a_y=a\cos \theta -g$

Here, the y-component of net acceleration of plane.

Rewrite the equation for $a_y$ by substituting relation for $a$.

$a_y=\left(\sqrt{a_x^2+g^2}\right)\cos \theta -g$ (II)

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.86\text{m}/{\text{s}}^{\text{2}}$ for $a_x$, and $\left(5.0°-2.0°\right)$ for $\theta$ in equation (I) to find $a_x$.

$\begin{array}{c}{a}_x=\left(\sqrt{{\left(0.86\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}^2}\right)\sin \left(5.0°-2.0°\right)\\ =\left(9.84\text{m}/{\text{s}}^{\text{2}}\right)\left(0.052\right)\\ =0.51\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.86\text{m}/{\text{s}}^{\text{2}}$ for $a_x$, and $\left(5.0°-2.0°\right)$ for $\theta$ in equation (II) to find $a_y$

$\begin{array}{c}{a}_y=\left(\sqrt{{\left(0.86\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}^2}\right)\cos \left(5.0°-2.0°\right)-9.8\text{m}/{\text{s}}^{\text{2}}\\ =\left(\left(9.84\text{m}/{\text{s}}^{\text{2}}\right)\left(0.99\right)\right)-9.8\text{m}/{\text{s}}^{\text{2}}\\ =0.02\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, the x-component of acceleration is $0.51\text{m}/{\text{s}}^{\text{2}}$ and the y-component of acceleration is $0.02\text{m}/{\text{s}}^{\text{2}}$.