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ssm 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 $\mathrm{m} / \mathrm{s}^{2}$, 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 $\mathrm{kg}$ of luggage were removed from car 2 and placed in (a) car 1 and (b) car 3? If the tension changes, specify whether it increases or decreases.

a) $\Delta T_A = 0\mathrm{N}$, $\Delta T_B = -4.7\mathrm{N}$, $\Delta T_C = 0\mathrm{N}$ b) $\Delta T_A = 0\mathrm{N}$, $\Delta T_B = 0\mathrm{N}$, $\Delta T_C = 4.7\mathrm{N}$

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University of Washington

Simon Fraser University

McMaster University

So this question. We have to be quite smart about the physics of the problem. Newton's second low tells us that the net force that acts on an object is given by the mass off that object times its acceleration show. The net force is responsible for accelerating that body. Keep that in mind in this question, we have three bars that are coupling the cars. So let me see that we have attention here in bar a disease. Attention A. We have attention here in Barbie is extension B and we have here. Attention. See, Inbar see off course. But aren't, uh, reaction forces off extensions. The other, like this diseases t pay. These is TV and this is T. C. Soon. In these diagram in red are the forces. The actions and orange are the reactions we will care particularly about the actions. Notice the following this horse T is what is making everything move. So th is responsible for making car 12 and three Move. Then we can say that Supposing that my reference frame points to the right. We can say that the eight is given by the mass off car 12 and three times the acceleration off the system. This is because TA is responsible for moving objects 12 and three. We can do the same for TV because the attention be is responsible for moving car street and three then then shall be is given by the mass off the second car president mass off the third card times the acceleration off the system and finally, PC is responsible for moving the last car. Therefore, TC is given by M Tree Times eight and noticed that I have used the same acceleration for every car. That is because they're moving together so they have the same acceleration, which is 0.12 meters per second squared. Knowing this, we can proceed to the question. So in all items we are removing 39 kilograms from Khartoum. But in the first item we are removing from Khartoum and transferring to car number one, how does attention changes among the cables? So attention eight is now given by M one plus Delta M one because we are adding more MasterCard number one plus entry plus Delta m. True because we had remove it sound mass from Khartoum Plus and Tree and Pre do not change. These is times acceleration. Noticed that Delta M one is equals two plus 39 kilograms. We are removing 39 kilograms from Karcher and adding 39 kilograms to car one. And then you can notice that there is a constellation because Delta M one is the question minus Delta M two. As a result, Ta Prime is just AM one plus choose plus and three times a two. It's the same your afford attention in Table A, though not change, no for two. Cable be we Have em True plus Delta M True Plus M three times eight and then disease M. Chu, minus 39 plus m three times eight. You can see that this is smaller down tension beat before meeting that luggage. Then we can calculate what is the difference between the regional tension on detention right now? So the variation in the tension be is given by the attention be no minors that things should be before and then his m true blows Delta M truth plus M Tree Times eight minus M. Chu plus M Tree Time State. You can see that a lot of terms we can. So so am true. Times eight will cancel on M three times. Eight will also cancel, So we're left with Delta AM Times A. So the change intention being is given by Delta M Truth Times eight and these is minus 39 times 0.12 which is approximately four 0.7 Utahans. And then we can calculate what is the tension See after moving the leverage that tension see is given by entry times eight. And it does not change because we're not messing with the mass off car number three. So summarizing we have ta being equal than it was before. TB decreasing by 4.7 Newton's there there was a minus sign here and finally TC being equal than it was before. Let me organize my board so I can continue shows in this question. The next item. What we do is that instead off moving the luggage to car number one, we move the luggage to card number three, so be moving to car number treat. Then Delta M three is equals a 39 kilograms. Let's see what happens with the tensions. The tension in car in table eight after is given by M one plus M two plus Delta M two plus M three plus Delta M Tree Times A. You can see that there is a cancellation between entry on entry, so we got the same tension as we had before. So the tension cable A. Those has changed again. Detention in cable be is given by M two plus Delta M True Plus M Tree plus Delta M Tree. Again. We have the cancellation between Delta m true on Delta M three. So the tension is the same again as it was before. And finally have detention. See which now is given by entry plus Delta entry times eight. You can see that he's is bigger down the tension. See, before meeting the luggage, we can calculate what is the variation? Intention Seat. This is TC Prime minus D. C, which is M three plus Delta Entry times A minus M treat times A. There is a cancellation that happens between M A tree and A and entry and a and really left with Delta TC. Mean equals two Delta M Tree Times eight, which is 39 times zero points 12 and these results in 4.7. Beautiful. Then we summarized the answer for the second item as the attention eight is the same. Attention B is the same, and only tension see changes, and it changes by 4.7 mu Tums, and this is the answer to the question.

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