At an airport, luggage is unloaded from a plane into the three cars of a luggage

Question

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 0.25 m /s2, and friction is negligible. The coupling bars have negligible mass.

The diagram shows a picture of a tractor pulling Car 1, 2 and 3each separated by a coupling bar, the first one attaching car 1 tothe tractor and titled coupling bar A and so on.



(a) By how much would the tension in each of the coupling bars A, B, and C change if 36 kg of luggage were removed from car 2 and placed in car 1? If the tension changes, specify whether it increases or decreases. (Specify a decrease in tension with a negative value and an increase in tension with a positive value. If there is no change, enter 0.)
?TA= N
?TB= N
?TC= N

(b) By how much would the tension in each of the coupling bars A, B, and C change if 36 kg of luggage were removed from car 1 and placed in car 3? If the tension changes, specify whether it increases or decreases. (Specify a decrease in tension with a negative value and an increase in tension with a positive value. If there is no change, enter 0.)
?TA= N
?TB= N
?TC= N

Explanation / Answer

a) Since TA is the bar which connects tractor with carriage A and since the masses behind it are not changing by just shuffling 36 kg from B to A therefore its tension of A will not change With same reasoning tension of B will change as masses behind it has changed but that of C will not since it doesn't see any change in mass. We use the equation Ta-Tb = .25*ma Tb-Tc = .25*mb Tc = .25*mc adding we get Ta = .25*(ma+mb+mc) substituting we get Tb = .25(mb+mc) Tc= .25*mc We can also write the equation for change of tension as dTa = .25*(dma + dmb +dmc) dTb = .25*(dmb +dmc) dTc = .25*(dmc) where dTi and dmi are changes in tension for rod i and changes in mass of carriage i Now coming to working out the answer we have 36 kg of luggage were removed from car 2 and placed in car 1 So, change in tension are dTa = .25*(+36-36+0) = 0 dTb = .25*(-36 +0) = -9N dTc = .25*0 = 0 b) 36 kg of luggage were removed from car 1 and placed in car 3 dTa = .25*(-36+0+36) = 0 dTb = .25*(0 +36) = 9N dTc = .25*36 = 9N

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