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/s}², 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 freebody diagrams for each.

What is the mass of the airplane?
a) 3000 , {kg}
b) 3300 , {kg}
c) 3600 , {kg}
d) 3900 , {kg}

Answer 1

To find the mass of the airplane, use Newton's second law of motion. Subtract the force resisting motion from the force exerted by the tractor and solve for mass. The mass of the airplane is approximately 3600 kg. Draw two separate free-body diagrams to solve each part of the problem.

Step-by-step explanation:

(a) To find the mass of the airplane, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). The total force exerted by the system resisting motion is 2400N, so we subtract this from the force exerted by the tractor (1.75 × 10^4 N) to get the net force. Using the net force and the given acceleration (0.150 m/s²), we can rearrange the formula to solve for mass (m = F/a). Substituting the values, we find that the mass of the airplane is approximately 3900 kg.

(b) To calculate the force exerted by the tractor on the airplane, we can use the equation F = ma. The force of friction experienced by the airplane is given as 2200N. Adding this force to the force resisting motion (2400N), we get the net force. Using the net force and the given acceleration (0.150 m/s²), we can solve for mass. Substituting the values, we find that the mass of the airplane is approximately 3600 kg.

(c) To solve each part, we can draw two separate free-body diagrams. For part (a), the system of interest includes the tractor and airplane. The forces acting on the tractor are the force exerted by the tractor on the pavement (1.75 × 10^4 N) and the force resisting motion (2400N). The force exerted by the tractor on the airplane can be represented as an arrow pointing backward. For part (b), the system of interest includes the tractor and airplane again, but this time we include the force of friction experienced by the airplane (2200N). The force exerted by the tractor on the airplane can also be represented as an arrow pointing backward.