At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car 1 and (b) car 3?

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Answer: I will assume car 1 is at the front, car 2 is in the middle, and 3 is at the back, and that coupling bar A is between the luggage carrier and car 1, B is between car 1 & 2, and C is between 2 & 3.

a) If 39 kg of luggage were removed from car 2 and placed in car 1:

A is pulling cars 1, 2, & 3 so:

The tension in A would not change, since the tension in A is pulling all three cars and the total mass of all three has not changed, just been moved a bit.

B is pulling cars 2 & 3 so:

The tension in B would DECREASE because the tension in B is pulling cars 2 & 3, and there are now 39kg less mass in car 2. The amount the tension would decrease by is:

F=ma

F= 39kg * 0.12m/s² = 4.68N

So the tension in B decreased by 4.86N

C is pulling only car 3, so:

The tension in C would not change since no mass was added or subtracted from car 3.

b) If 39 kg of luggage were removed from car 2 and placed in car 3:

A is pulling cars 1, 2, & 3 so:

The tension in A would not change, since the tension in A is pulling all three cars and the total mass of all three has not changed, just been moved around.

B is pulling cars 2 & 3 so:

The tension in B would not change, since the tension in B is pulling cars 2 & 3 and the total mass of 2 & 3 has not changed, just been moved around.

C is pulling only car 3, so:

The tension in C would INCREASE since 39kg mass was added to car 3. The tension would increse by:

F=ma

F= 39kg * 0.12m/s² = 4.68N

So the tension in C increased by 4.86N

I assume there was a diagram that actually showed the positions of the cars. Hopefully my assumptions about their positions are right.