Chapter 9.2, Problem 31SSC

(a)

To determine

The definition of two-cart system.

The speed of the cart C at the bottom of the incline.

Answer to Problem 31SSC

The velocity of the cart C at the bottom of the incline is $v_{\text{f}}=3.13\text{m/s}$ .

Explanation of Solution

Given info:

The weight of the cart at the top of the ramp is $F_{\text{C}}=24.5\text{N}$ .

The weight of the cart at the bottom of the ramp is $F_{\text{D}}=36.8\text{N}$ .

The angle made by the ramp with the horizontal is $\theta =30°$ .

The length of the ramp is $s=1\text{m}$ .

Calculation:

Show the system of the carts as shown in the Figure below:

Refer the Figure,

Two Cart System:

Consider the cart at the top of the ramp is denoted by cart C.

Consider the cart at the bottom of the ramp is denoted by cart D.

The cart D is initially at rest.

The cart C which is initially at rest, start moving down the ramp.

Find the acceleration of the cart C using the relation:

$\begin{array}{l}{F}_{\text{N}}\sin \theta =ma\\ mg\sin \theta =ma\\ a=g\sin \theta \end{array}$

Show the relation between the velocity and acceleration of the cart C as follows:

$v_{\text{f}}^2=v_{\text{i}}^2+2as$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ is the initial and final velocity of the cart, $a$ is the acceleration of the cart, and $s$ is the distance travelled by the cart.

The initial velocity of the cart C is $v_{\text{i}}=0$ .

Find the final velocity of the cart C as follows:

$\begin{array}{l}{v}_{\text{f}}^2=0^2+2as\\ v_{\text{f}}=\sqrt{2gs\sin \theta }\\ v_{\text{f}}=\sqrt{2\times \left(9.80{\text{m/s}}^2\right)\left(1.00\text{m}\right)\sin \left(30°\right)}\\ v_{\text{f}}=3.13\text{m/s}\end{array}$

Hence, the velocity of the cart C at the bottom of the incline is $v_{\text{f}}=3.13\text{m/s}$ .

Conclusion:

Thus, the velocity of the cart C at the bottom of the incline is $v_{\text{f}}=3.13\text{m/s}$ .

(b)

To determine

The initial velocity of the carts which stick together.

Answer to Problem 31SSC

The velocity of the carts which are stick together is $v=1.25\text{m/s}$ .

Explanation of Solution

Given info:

The weight of the cart at the top of the ramp is $F_{\text{C}}=24.5\text{N}$ .

The weight of the cart at the bottom of the ramp is $F_{\text{D}}=36.8\text{N}$ .

The angle made by the ramp with the horizontal is $\theta =30°$ .

The length of the ramp is $s=1\text{m}$ .

Calculation:

Consider the acceleration due to gravity as $g=9.80{\text{m/s}}^2$ .

The initial velocity of the cart C is denoted by $v_{\text{Ci}}=3.13\text{m/s}$ .

The initial velocity of the cart D is denoted by $v_{\text{Di}}=0\text{m/s}$ .

The mass of the cart C is denoted by $m_{\text{C}}$ .

The mass of the cart D is denoted by $m_{\text{D}}$ .

Find the velocity of the cart which stick together, using the relation:

$m_{\text{C}}{v}_{\text{Ci}}+m_{\text{D}}{v}_{\text{Di}}=\left(m_{\text{C}}+m_{\text{D}}\right)v$

Multiply both sides of the Equation by the acceleration due to gravity $g$ .

  $\begin{array}{l}{m}_{\text{C}}g{v}_{\text{Ci}}+m_{\text{D}}g{v}_{\text{Di}}=\left(m_{\text{C}}g+m_{\text{D}}g\right)v\\ F_{\text{C}}{v}_{\text{Ci}}+F_{\text{D}}{v}_{\text{Di}}=\left(F_{\text{C}}+F_{\text{D}}\right)v\\ v=\frac{\left(24.5\text{N}\right)\left(3.13\text{m/s}\right)+\left(36.8\text{N}\right)\left(0\text{m/s}\right)}{24.5\text{N}+36.8\text{N}}\\ v=\frac{\left(24.5\text{N}\right)\left(3.13\text{m/s}\right)}{24.5\text{N}+36.8\text{N}}\\ v=1.25\text{m/s}\end{array}$

Conclusion:

Thus, the velocity of the cart which are stick together is $v=1.25\text{m/s}$ .