(II) One 3.2-kg paint bucket is hanging by a massless cord fro...

Question

(II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4–41.

(b) If the two buckets are pulled upward with an acceleration of 1.45m/s² by the upper cord, calculate the tension in each cord.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice prom together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. 13.3 kg lantern hangs from a weightless string attached to another 3.3 kg lantern which is also suspended by a weightless string. Calculate the tension in each string if the two lanterns are pulled upward with an acceleration of 1.55 m per second squared by the upper string. So that's our end goal. So ultimately, we're asked to figure out what the tension in each string is if the two lanterns are pulled upward with an acceleration of 1.55 m per second squared by the upper string. So that will be our final answers when we calculate the tension in each string. So that means we're solving for two separate answers. And once we find those two separate answers, that means we've officially solved for this problem. Awesome. So we're also provided a figure by the pro itself. And as we can see, we have our string one, which is connected to our first lantern and then below our second. So below our first lantern is our second lantern also connected by a string. Awesome. So with that in mind, now that we have our problem properly visualized, let's look at our multiple choice answers to see which answer pair might be our final answer. And let's also note that the tension is for the units, for the tension, for both strings are all in units of newtons. So for a, it's 37 and 75 B is 30 70 C is 25 and 65 and D is 2060. Awesome. So first off, let us create free body diagrams for each of the lanterns to better visualize what forces are at work here. So I've gone ahead and found a detailed diagram of our lanterns to help visualize our free body diagrams. As you can see on the far left of the diagram that I just shared with you all, we have the force and the tension. So, so basically F denotes the force, and T denotes the tension. So at the force and the 12th tension. So T two, so the force of T two. So this is considering our bottom lantern here, we have the force of tension T one. So we have the force of the first tension in the first string for the first lantern. So looking at our, looking to the left left here, far left, we have our first lantern and then our second lantern. So the second lantern is not only experiencing the tension force of its string in between both lanterns, but it's also experiencing the tension in the first string up above the first lantern. It's also experiencing the mass of the lanterns themselves plus gravity, pulling everything back down to earth. So we have mass pulling down on the string and we also have gravity, pushing or pulling, I should say not pushing, pulling down on the string. So similarly for focusing on our first lantern here to the right of our diagram, we have the force and of T one. And also like before, we have not only the mass of the lantern pulling it down towards the ground, we also have gravity pulling it down towards the ground. OK. So with that in mind, let us note that lantern one will be the lower lantern and lantern two will be the upper lantern. Let us also note that ft one is the tension in the string connecting the upper and lower lanterns and FT two is the tension in the string supporting the upper lantern. Now, we can recall and apply Newton's second law for each of the lanterns, but make sure we are using a nonzero acceleration. And let us also note that the sum of F one and the sum of F two are the sum of the forces on lantern one and lantern two. Now, if we apply Newton's second law, the net force for the sum of F one will be the tension in the string minus the gravitational force on lantern one, which we can write as the sum of F one is equal to FT one. So FT one FT one minus mass multiplied by gravity is equal to mass multiplied by acceleration. OK. So we can go even further and write that FT one is equal to M multiplied by G plus M multiplied by A. So now we can plug in all of our known values to solve for FT one. So FT one is equal to 3.3 kg because that's the mass of our lantern or lantern. Since they're both 3.3 kg multiplied by gravity, which numerically gravity is 9.8 m per second squared plus 1.5 5 m per second squared means that's the acceleration that's provided to us in the prong itself. So when we plug that into our calculator, we will find that FT one is equal to 37 newtons. And this is when we round to the nearest hole number. And that is our first answer. Now we can start to solve for FT two. So for lantern two, considering if the net force, the sum of F two is the difference in tension between FT two and FT one. Accounting for the acceleration, we could write that the sum of oops I'm sorry about that. OK. So the sum of drum roll, F two is equal to FT two minus FT one minus M multiplied by G is all equal to M multiplied by A. We can now take this equation and we can rearrange it to isolate and solve for FT two. So when we do that, we will find that FT two is equal to FT one plus M multiplied by G plus M multiplied by A which is all equal to two multiplied by FT one because FT one and the mass multiplied by the gravity plus mass multiplied by the acceleration will represent the same amount of force required to support one lantern. Since FT two has to support both lanterns. FT one must be doubled as a result of this fact. Therefore, we could write that FT two is equal to two multiplied by FT one, which is equal to when we plug in our known variables, two multiplied by 37 which we have to pause here for a moment here. So earlier, when we saw for the force of T one, we rounded to the nearest whole number which was 37 newtons. But it's important to note that when you initially plug this equation into your calculator, you'll get 37.45 newtons. And we are gonna use this value to plug in the sulfur FT two. So 37.45 newtons. So when we multiply two by 37.45 newtons, we will get 74.9 newtons which rounding to the nearest whole number. Because as you can see all of our multiple choice answers are all rounded to the nearest whole number. That will mean that our answer is approximately 75 Newtons. And that is our final answer. Hooray, we did it so to quickly recap here, 37 Newtons is our FT one value and 75 newtons is equal to our FT two value. Awesome. Therefore, the tensions FT one and FT two are determined by taking the sum of the forces on each lantern. So the sum of F one and the sum of F two and they will be consistent with Newton's second law which is visualized in our free body diagram pictures. So therefore without a doubt, multiple choice answer letter A has to be our final answer. So A is 37 Newtons and 75 Newtons. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

(II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4–41.
<IMAGE>
(b) If the two buckets are pulled upward with an acceleration of 1.45m/s² by the upper cord, calculate the tension in each cord.

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Forces in Connected Systems of Objects practice set

Struggling with Physics?

(II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4–41.<IMAGE>(b) If the two buckets are pulled upward with an acceleration of 1.45m/s² by the upper cord, calculate the tension in each cord.

Watch next

Forces in Connected Systems of Objects practice set

(II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4–41.
<IMAGE>
(b) If the two buckets are pulled upward with an acceleration of 1.45m/s² by the upper cord, calculate the tension in each cord.