Chapter 4, Problem 48P

(a)

To determine

The relation between the components of velocity.

Answer to Problem 48P

The relation between the components of velocity is $v_x=u{v}_y$.

Explanation of Solution

Write the equation for the position of the glider.

  $x={\left(z^2-h_0^2\right)}^{1/2}$        (I)

Here, $x$ is the position of the glider, $z$ is the coordinate, $h_0$ is the height.

Write the expression for the x-component of velocity.

$v_x=\frac{dx}{dt}$        (II)

Here, $v_x$ is the x-component of velocity and $\left(dx/dt\right)$ is the rate of change of position.

Write the expression for the y-component of velocity.

$v_y=\frac{dz}{dt}$        (III)

Here, $v_y$ is the y-component of velocity and $\left(dz/dt\right)$ is the rate of change of position.

Conclusion:

Substitute ${\left(z^2-h_0^2\right)}^{1/2}$ for $x$, $v_y$ for $\frac{dz}{dt}$ in Equation (II) to find the relation.

  $\begin{array}{c}{v}_x=\frac{d\left[{\left(z^2-h_0^2\right)}^{1/2}\right]}{dt}\\ =\frac{1}{2}{\left(z^2-h_0^2\right)}^{-1/2}\left(2z\right)\left(\frac{dz}{dt}\right)\\ =z{\left(z^2-h_0^2\right)}^{-1/2}{v}_y\\ =u{v}_y\end{array}$

Thus, the relation between the components of velocity is $v_x=u{v}_y$.

(b)

To determine

The relation between the components of the acceleration.

Answer to Problem 48P

The relation between the components of the acceleration is $\underset{\_}{a_x=u{a}_y}$.

Explanation of Solution

Write the equation for the x-component of acceleration.

  $a_x=\frac{d{v}_x}{dt}$        (IV)

Here, $a_x$ is the x-component of acceleration.

Write the equation for the y-component of acceleration.

  $a_y=\frac{d{v}_y}{dt}$        (V)

Here, $a_y$ is the y-component of acceleration.

Since, the glider release from rest, $v_y=0$.

Conclusion:

Substitute $u{v}_y$ for $v_x$, $a_y$ for $\left(d{v}_y/dt\right)$, $0$ for $v_y$ in Equation (IV) to find the relation.

  $\begin{array}{c}{a}_x=\frac{d\left(u{v}_y\right)}{dt}\\ =u\left(\frac{d{v}_y}{dt}\right)+v_y\left(\frac{du}{dt}\right)\\ =u{a}_y\end{array}$

Thus, the relation between the components of the acceleration is $\underset{\_}{a_x=u{a}_y}$.

(c)

To determine

The tension in the string.

Answer to Problem 48P

The tension in the string is $\underset{\_}{3.56\text{N}}$.

Explanation of Solution

Write the equation of motion for the counterweight.

  $T-m_{1}g=-m_1{a}_y$        (VI)

Here, $T$ is the tension, $m_1$ is the mass and $g$ is the gravitational acceleration.

Write the expression for the coordinate.

$z=\frac{\sin \theta }{h_0}$        (VII)

Here, $\theta$ is angle.

Write the expression for the position of glider.

$\begin{array}{c}u={\left(z^2-h_0^2\right)}^{-1/2}\\ ={\left({\left(\frac{\sin \theta }{h_0}\right)}^2-h_0^2\right)}^{-1/2}\end{array}$        (VIII)

Write the equation of motion for the glider by using equation (VI) and (VIII).

  $\begin{array}{c}T\cos \theta =m_2{a}_x\\ =m_{2}u{a}_y\\ =m_2{\left[{\left(\frac{\sin \theta }{h_0}\right)}^2-h_0^2\right]}^{-1/2}\left[\frac{T-m_{1}g}{\left(-m_1\right)}\right]\end{array}$        (IX)

Conclusion:

Substitute $1.00\text{kg}$ for $m_2$, $30°$ for $\theta$, $80\text{cm}$ for $h_0$, $0.500\text{kg}$ for $m_1$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ in Equation (IX) to $T$.

  $\begin{array}{c}T\cos \left(30\right)=\left\{\begin{array}{l}\left(1.00\text{kg}\right){\left[{\left(\frac{\sin \left(30\right)}{\left[\left(80\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right)}^2-{\left[\left(80\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2\right]}^{-1/2}\\ \left[\frac{T-\left(0.500\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)}{\left(-\left(0.500\text{kg}\right)\right)}\right]\end{array}\right\}\\ T\left(0.866\right)=-2.31T+11.3\text{N}\\ T=3.56\text{N}\end{array}$

Thus, the tension in the string is $\underset{\_}{3.56\text{N}}$.