Math Problem Statement

Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) The 1800 kg tractor exerts a force of 1.95 104 N backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.100 m/s2, what is the mass of the airplane in kilograms? (b) Calculate the force (in N) exerted by the tractor on the airplane, assuming 2000 N of the friction is experienced by the airplane.

Solution

Let's solve the problem step by step.

Part (a) - Finding the mass of the airplane:

We are given:

  • The force exerted by the tractor on the pavement: Ftractor=1.95×104NF_{\text{tractor}} = 1.95 \times 10^4 \, \text{N}.
  • The total resistive forces acting on the system: Fresist=2400NF_{\text{resist}} = 2400 \, \text{N}.
  • The acceleration of the system: a=0.100m/s2a = 0.100 \, \text{m/s}^2.
  • The mass of the tractor: mtractor=1800kgm_{\text{tractor}} = 1800 \, \text{kg}.

Now, to find the mass of the airplane, we'll apply Newton's second law to the system as a whole. The net force FnetF_{\text{net}} acting on the system is given by the difference between the force exerted by the tractor and the resistive forces: Fnet=FtractorFresist=1.95×104N2400N=1.71×104NF_{\text{net}} = F_{\text{tractor}} - F_{\text{resist}} = 1.95 \times 10^4 \, \text{N} - 2400 \, \text{N} = 1.71 \times 10^4 \, \text{N}

From Newton's second law Fnet=(mtractor+mairplane)×aF_{\text{net}} = (m_{\text{tractor}} + m_{\text{airplane}}) \times a, we can solve for the mass of the airplane mairplanem_{\text{airplane}}:

1.71×104N=(1800kg+mairplane)×0.100m/s21.71 \times 10^4 \, \text{N} = (1800 \, \text{kg} + m_{\text{airplane}}) \times 0.100 \, \text{m/s}^2

Solving for mairplanem_{\text{airplane}}:

mairplane=1.71×104N0.100m/s21800kgm_{\text{airplane}} = \frac{1.71 \times 10^4 \, \text{N}}{0.100 \, \text{m/s}^2} - 1800 \, \text{kg}

mairplane=171000kg1800kgm_{\text{airplane}} = 171000 \, \text{kg} - 1800 \, \text{kg}

mairplane=169200kgm_{\text{airplane}} = 169200 \, \text{kg}

So, the mass of the airplane is 169200 kg.

Part (b) - Finding the force exerted by the tractor on the airplane:

We are given:

  • The friction experienced by the airplane: Ffriction=2000NF_{\text{friction}} = 2000 \, \text{N}.

We need to find the force exerted by the tractor on the airplane. Using Newton's second law for the airplane alone:

Ftractor on airplaneFfriction=mairplane×aF_{\text{tractor on airplane}} - F_{\text{friction}} = m_{\text{airplane}} \times a

Substituting the values we know:

Ftractor on airplane2000N=169200kg×0.100m/s2F_{\text{tractor on airplane}} - 2000 \, \text{N} = 169200 \, \text{kg} \times 0.100 \, \text{m/s}^2

Ftractor on airplane2000N=16920NF_{\text{tractor on airplane}} - 2000 \, \text{N} = 16920 \, \text{N}

Solving for Ftractor on airplaneF_{\text{tractor on airplane}}:

Ftractor on airplane=16920N+2000NF_{\text{tractor on airplane}} = 16920 \, \text{N} + 2000 \, \text{N}

Ftractor on airplane=18920NF_{\text{tractor on airplane}} = 18920 \, \text{N}

Thus, the force exerted by the tractor on the airplane is 18920 N.


Let me know if you need any details or have further questions!

Here are 5 questions that expand upon this topic:

  1. How would the force change if the resistive forces acting on the system increased?
  2. What happens to the acceleration if the mass of the airplane increases while keeping the tractor's force constant?
  3. How does the system behave if the friction experienced by the airplane doubles?
  4. How would the tractor’s force change if the acceleration increased to 0.200 m/s²?
  5. What is the significance of knowing the force of friction in systems like this?

Tip: Always ensure to subtract resistive forces when calculating net force in problems involving motion!

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Force
Friction
Mass
Acceleration

Formulas

F_net = F_tractor - F_resist
F_net = (m_tractor + m_airplane) × a
F_tractor on airplane = F_airplane × a + F_friction

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12