Question

In April 1974, John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed $\mathbf{700}\mathbf{}\mathbf{kN}$(about$\mathbf{80}\mathbf{}\mathit{t}\mathit{o}\mathit{n}\mathit{s}$). Assume that he pulled with a constant force that was 2.5 times his body weight, at an upward angle$\mathit{\theta }$of${\mathbf{30}}^{\mathbf{°}}$from the horizontal. His mass was$\mathbf{80}\mathbf{}\mathbf{kg}$and he moved the cars by$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$. Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.

Short answer

The speed of the car at the end of the pull is$0.22\mathrm{m}/\mathrm{s}$ .

Step-by-step solution

The given data

• Total weight of the cars, $W_{cars}=700\mathrm{kN}$.
• Pulling force exerted by John.$=2.5\times John\text{'}sweight.$
• Mass of John, $m=80\mathrm{kg}$
• Upward angle of motion, $\theta ={30}^{°}.$
• Distance of pull by John,$\times =1.0\mathrm{m}.$

Understanding the concept of kinematics

Using the free body diagram and Newton’s 2nd law of motion we can find the acceleration of the cars. As we know the displacement of the cars and the acceleration of the car, we can find the speed of the cars.

Formulae:

Force according to Newton’s second law,$F=ma$ (i)

The third kinematic equation of motion, ${v_f}^2={v_i}^2+2ax$ (ii)

Calculation of speed of car at the end of pull

Free body diagram

$\underset{}{\sum F_{\times }}=0$

$F\cos \theta =ma....................\left(a\right)$

role="math" localid="1657000764711" $$\begin{gathered} (\because \mathrm{From}\mathrm{diagram},\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{the}\mathrm{force}\mathrm{by}\mathrm{acceleration}\mathrm{is} \\ \mathrm{balanced}\mathrm{by}\mathrm{cosine}\mathrm{component}\mathrm{of}\mathrm{force}\mathrm{and}\mathrm{from}\mathrm{equation}(\mathrm{i}\left)\right) \end{gathered}$$

As we have been given,

$$\begin{gathered} F_{\mathrm{pulling}\mathrm{fo}\mathrm{roe}\mathrm{exerted}\mathrm{by}\mathrm{john}}=F \\ =2.5\times \mathrm{weight}\mathrm{of}\mathrm{john} \\ =2.5\times mg \\ =2.5\times 80\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ =1960\mathrm{N} \end{gathered}$$

From equation (a) and the given values, the acceleration is given as:

role="math" localid="1657000847621" $$\begin{gathered} 1960\mathrm{N}\times \cos {30}^{°}=\frac{7\times {10}^5\mathrm{N}}{g}\times \mathrm{a} \\ \mathrm{a}=\frac{1695.4\times 9.8}{7\times {10}^5} \\ =23.7\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From equation (ii) and the given values, the final speed of the car is given as:

$$\begin{gathered} v_f^2=0+23.7\times {10}^{-3}\times 2\times 1.0 \\ {v}_f=0.219\mathrm{m}/\mathrm{s} \\ =0.22\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of car is$0.22\mathrm{m}/\mathrm{s}$ .