Question

Question: A flat cushion of mass $\mathit{m}$is released from rest at the corner of the roof of a building, at height $\mathit{h}$. A wind blowing along the side of the building exerts a constant horizontal force of magnitude $\mathit{F}$on the cushion as it drops as shown in Figure P5.91. The air exerts no vertical force. (a) Show that the path of the cushion is a straight line. (b) Does the cushion fall with constant velocity? Explain. (c) If $\mathit{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{kg}$, $\mathit{h}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$, and $\mathit{F}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{N}$, how far from the building will the cushion hit the level ground? What If? (d) If the cushion is thrown downward with a nonzero speed at the top of the building, what will be the shape of its trajectory? Explain.

Short answer

(a)The path of the cushion is straight line.

(b) The cushion velocity does not fall constant velocity.

(c) The maximum horizontal displacement is $1.63\mathrm{m}$.

(d) The shape of trajectory is curved downwards.

Step-by-step solution

Writing the given data from the question

Mass of the cushion is $m$.

Height of the Building is $h$.

Magnitude of the constant horizontal force is $F$.

Determining the required formulas

The expression to calculate the force on the object is given as follows.

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, $\mathit{F}$is the working acting on object, $\mathit{m}$is the mass of the object and $\mathit{a}$is the acceleration of the object.

The expression to calculate the vertical distance is given as follows.

$\mathit{y}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$

Here,$\mathit{u}$is the velocity, $\mathit{t}$is the time, and${\mathit{a}}_{\mathbf{y}}$is the acceleration in the vertical direction.

 Step 3: (a) Showing that the path of the cushion is a straight line

Let assume the displacement horizontal displacement is $x$and vertical displacement is $y$after the time $t$.

The horizontal acceleration of the cushion is given by

$a_x=\frac{F}{m}$

The horizontal displacement is given by

$$\begin{gathered} x=\frac{1}{2}{a}_x{t}^2 \\ 2x=a_x{t}^2 \\ {t}^2=\frac{2x}{a_x} \\ t=\sqrt{\frac{2x}{a_x}} \end{gathered}$$

Substitute $\frac{F}{m}$for $a_x$into the above equation.

$\begin{array}{c}t=\sqrt{\frac{2x}{{\displaystyle \frac{F}{m}}}}\\ =\sqrt{\frac{2xm}{F}}\end{array}$

The vertical acceleration of the cushion is given by

$a_y=g$

The vertical displacement is given by

$y=\frac{1}{2}{a}_y{t}^2$

Substitute $g$for $a_y$and $\sqrt{\frac{2xm}{F}}$for $t$into the above equation.

$\begin{array}{rcl}y& =& \frac{1}{2}g{\left(\sqrt{\frac{2xm}{F}}\right)}^2\\ & =& \frac{1}{2}g\times \frac{2xm}{F}\\ & =& g\times \frac{xm}{F}\\ & =& \left(\frac{gm}{F}\right)x\end{array}$

Let us assume that $g$,$m$and $f$are the constants which are represented by$k$.

$y=kx$

Therefore, the above equation is a straight-line equation.

Hence, the path of the cushion is a straight line.

(b) Determining if the cushion fall with the constant velocity

Since the cushion is under the acceleration in horizontal and vertical direction, changes with the time. Therefore, resultant velocity $v$is given as,

$v=\sqrt{v_x^2+v_y^2}$

Here, $v_x$and $v_y$are the horizontal and vertical velocities of the cushion.

Hence, the cushion velocity does not fall constant velocity.

(c) Calculating the maximum horizontal displacement

Consider the given data as below.

Mass of the cushion, $m=1.20\mathrm{kg}$

Height of the building, $h=8\mathrm{m}$

Horizontal wind force, $F=2.40\mathrm{N}$

The maximum horizontal displacement is obtained when $y=h$and can be calculated as

role="math" localid="1663663650553" $$\begin{gathered} \mathrm{R}=\frac{\mathrm{Fh}}{\mathrm{gm}} \\ \mathrm{R}=\frac{24\mathrm{N}\times 8\mathrm{m}}{9.8\mathrm{m}/{\mathrm{s}}^2\times 1.2\mathrm{kg}} \\ \mathrm{R}=1.63\mathrm{m} \end{gathered}$$

Hence, the maximum horizontal displacement is $1.63\mathrm{m}$.

(d) Determining the shape of trajectory

$t$Let assume cushion is thrown with the velocity $u$.

The vertical displacement is given by

$y=ut+\frac{1}{2}{a}_y{t}^2$

Substitute $g$for $t$and $\sqrt{\frac{2xm}{F}}$for into the above equation.

$\begin{array}{rcl}y& =& u\sqrt{\frac{2xm}{F}}+\frac{1}{2}g\times {\left(\sqrt{\frac{2xm}{F}}\right)}^2\\ & =& u\sqrt{\frac{2xm}{F}}+\frac{1}{2}g\times \frac{2xm}{F}\\ & =& \sqrt{\frac{m}{F}}\left(\sqrt{2u}\sqrt{x}+\sqrt{\frac{m}{F}}gx\right)\\ & & \end{array}$

The above equation contains both the terms $x$ and $\sqrt{x}$, therefore, the shape of trajectory is curved downwards.

Hence, the shape of trajectory is curved downwards.