Question

name third law and eliminating internal forces
scenario
a train engine pulls a train with three cars. each car has the mass shown. suppose that the cars are connected by metal bars with the tensions indicated in the diagram. the engine accelerates at a rate of 2 m/s². assume that the cars travel on bearings with negligible friction.
part a: using representations
the dots below represent the three train cars. draw free - body diagrams showing and labeling the forces (not components) exerted on each car. draw the relative lengths of all vectors to reflect the relative magnitudes of all the forces. each force must be represented by a distinct arrow starting on and pointing away from the dot. for each diagram, write an equation that relates the horizontal forces in the diagram to acceleration.
forces on the 3,000 kg car
forces on the 2,000 kg car
forces on the 1,000 kg car
part b: the dots below represent three different systems. draw free - body diagrams showing and labeling the forces (not components) exerted on each system. draw the relative lengths of all the vectors to reflect the relative magnitudes of all the forces. each force must be represented by a distinct arrow starting on and pointing away from the dot. for each diagram, write an equation that relates the forces in the diagram to acceleration.
forces on the system of the 2,000 kg and 3,000 kg cars
forces on the system of the 2,000 kg and 1,000 kg cars
forces on system of the 3,000 kg, 2,000 kg, and 1,000 kg cars

Explanation:

Step1: Identify forces on each car

For a single - car system, the only horizontal force is the tension pulling it forward. According to Newton's second law $F = ma$, where $F$ is the net force, $m$ is the mass, and $a$ is the acceleration.

Step2: Analyze forces on combined - car systems

For combined - car systems, the net force acting on the system is what causes the acceleration of the combined mass. Again, using $F = ma$, where $F$ is the tension pulling the combined cars and $m$ is the total mass of the combined cars.

Forces on the 3,000 kg car

• Free - body diagram: One horizontal arrow (tension $F_3$) pointing to the right starting from the dot.
• Equation: $F_3=3000\times a$, since $a = 2m/s^2$, $F_3 = 3000\times2=6000N$

Forces on the 2,000 kg car

• Free - body diagram: One horizontal arrow (tension $F_2$) pointing to the right starting from the dot.
• Equation: $F_2=2000\times a$, so $F_2=2000\times2 = 4000N$

Forces on the 1,000 kg car

• Free - body diagram: One horizontal arrow (tension $F_1$) pointing to the right starting from the dot.
• Equation: $F_1=1000\times a$, so $F_1=1000\times2=2000N$

Forces on the system of the 2,000 kg and 3,000 kg cars

• Free - body diagram: One horizontal arrow (tension pulling both cars) pointing to the right starting from the dot.
• Total mass $m_{total}=2000 + 3000=5000kg$
• Equation: $F=5000\times a$, so $F = 5000\times2=10000N$

Forces on the system of the 2,000 kg and 1,000 kg cars

• Free - body diagram: One horizontal arrow (tension pulling both cars) pointing to the right starting from the dot.
• Total mass $m_{total}=2000 + 1000=3000kg$
• Equation: $F=3000\times a$, so $F = 3000\times2=6000N$

Forces on system of the 3,000 kg, 2,000 kg, and 1,000 kg cars

• Free - body diagram: One horizontal arrow (tension pulling all three cars) pointing to the right starting from the dot.
• Total mass $m_{total}=3000+2000 + 1000=6000kg$
• Equation: $F=6000\times a$, so $F = 6000\times2=12000N$

Answer:

Forces on the 3,000 kg car: Free - body diagram has one right - pointing arrow labeled $F_3$, equation $F_3 = 6000N$
Forces on the 2,000 kg car: Free - body diagram has one right - pointing arrow labeled $F_2$, equation $F_2 = 4000N$
Forces on the 1,000 kg car: Free - body diagram has one right - pointing arrow labeled $F_1$, equation $F_1 = 2000N$
Forces on the system of the 2,000 kg and 3,000 kg cars: Free - body diagram has one right - pointing arrow, equation $F = 10000N$
Forces on the system of the 2,000 kg and 1,000 kg cars: Free - body diagram has one right - pointing arrow, equation $F = 6000N$
Forces on system of the 3,000 kg, 2,000 kg, and 1,000 kg cars: Free - body diagram has one right - pointing arrow, equation $F = 12000N$