Q4.6-28PE - College Physics (Urone)

Question

Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. 

(a) An 1800-kg tractor exerts a force of 1.75×10⁴ N backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s², what is the mass of the airplane?

(b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. 

(c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

Step-by-Step Solution

Answer

(a) The mass of the airplane is 98866.67 kg.

(b) The force exerted by the tractor is 17030 N.

Step 1: Given Data

• Mass of the tractor = 1800 kg.
• Force exerted by the tractor = 1.75×10⁴ N.
• Resisting forces = 2400 N.
• Acceleration = 0.150 m/s².
• the friction is experienced by the airplane = 2200 N.

Step 2: (a) Determine the mass of the airplane

Apply Newton’s second law of motion as:

 

$\begin{array}{c}{\mathit{F}}_{\mathit{n}\mathit{e}\mathit{t}}\mathbf{=}\mathit{M}\mathit{a}\\ \mathit{F}\mathbf{-}\mathit{f}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{a}}\mathbf{+}{\mathbf{m}}_{\mathbf{t}}\mathbf{)}\mathit{a}\\ {\mathit{m}}_{\mathit{a}}\mathbf{=}\frac{\mathit{F}\mathbf{-}\mathit{f}}{\mathit{a}}\mathbf{-}{\mathit{m}}_{\mathit{t}}\end{array}$………………… (i)

Here, F_net is the net force acting on the system, m_a is the mass of the airplane, m_t is the mass of the tractor, a is the acceleration, F is the force exerted by the tractor, and f is the friction force.

 

Substitute 1800 kg for m_t, 2400N for f, 1.75x10⁴ N for F, and 0.150 m/s² for a in equation (1), and we get,

 

$\begin{array}{c}{m}_a=\frac{17500\text{\hspace{0.33em}N}-2400\text{\hspace{0.33em}N}}{0.150{\text{\hspace{0.33em}m/s}}^{\text{2}}}-1800\text{\hspace{0.33em}kg}\\ =\frac{15100\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^{\text{2}}}{0.150{\text{\hspace{0.33em}m/s}}^{\text{2}}}-1800\text{\hspace{0.33em}kg}\\ =100666.67\text{\hspace{0.33em}kg}-1800\text{\hspace{0.33em}kg}\\ =98866.67\text{\hspace{0.33em}kg}\end{array}$

 

Hence, the mass of the airplane is 98866.67 kg.

Step 3: (b) Determine the force exerted by the tractor, considering friction

Apply Newton’s second law of motion as:

 

$\mathit{F}\mathbf{\text{'}}\mathbf{-}\mathit{f}\mathbf{\text{'}}\mathbf{=}{\mathit{m}}_{\mathbf{a}}\mathit{a}$

 

Here, F’ is the force exerted by the tractor, and f’ is the friction force.

 

Substitute 2200N for f’, 98866.67 kg for m_a, and 0.150 m/s² for a in the above expression, and we get,

 

$\begin{array}{c}F\text{'}-2200\text{\hspace{0.33em}N}=98866.67\text{\hspace{0.33em}kg}\times 0.150{\text{\hspace{0.33em}m/s}}^{\text{2}}\\ F\text{'}=14830\text{\hspace{0.33em}N+2200\hspace{0.33em}N}\\ F\text{'}=\text{17030\hspace{0.33em}N}\end{array}$

 

Hence, the force exerted by the tractor is 17030 N.

Step 4: (c) The free-body diagrams for the above two cases