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Consider the masses shown_ Assume the table is frictionless (and very tall) and the pulley is massless and frictionless. Find the acceleration of each mass 20kg 10k...

Question

Consider the masses shown_ Assume the table is frictionless (and very tall) and the pulley is massless and frictionless. Find the acceleration of each mass 20kg 10kg (b) Each of the two strings will have its own tension. Find these twO tensions . Show all steps in your solution'30kg

Consider the masses shown_ Assume the table is frictionless (and very tall) and the pulley is massless and frictionless. Find the acceleration of each mass 20kg 10kg (b) Each of the two strings will have its own tension. Find these twO tensions . Show all steps in your solution' 30kg



Answers

Two bodies of mass $4 \mathrm{~kg}$ and $6 \mathrm{~kg}$ are attached to the ends of a string passing over a pulley (see Fig. 7,311 ). The 4 $\mathrm{kg}$ mass is attached to the table top by another string. The tension in this string $T_{1}$ is equal to (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ) a. $20 \mathrm{~N}$ b. $25 \mathrm{~N}$ c. $10.6 \mathrm{~N}$ d. $10 \mathrm{~N}$

Uh this question we again that our two blocks are connected by a massless rope has shown the mass of the block on the table is four kg. And the hanging mouse and one is one kg. And the table and the pulley are friction as everything is frictionless. So we have to find the acceleration of the system. So let's let's draw the fPd. So here we have attention here we have another tension. Here we have the full strength of gravity. So it's gonna be one times G, which is just G. And this is a four kg. So this is the reaction force. This is the acceleration due to gravity for G. So the system is going right by then definitely it should be going down by as well. So the acceleration has got to be t is equal to anyone ever and one is four. So it's going to be equal to four images equation one. And uh the other equation is going to be four G minus T is equal to M two times. So it does one time say which is just a. Which is equation too. We are interested in finding, finding the accelerations. If we add this up, we have G is equal to five, A G is equal to five, which means that acceleration is equal to G over five And G is nothing but 9.8 m/s. So this is nine it over five, This is nine point 8/5. So that's 9.8 or five inches 1.96. So this is 1.96 m per second square. Uh This is the acceleration. Uh Then we have to find the tension of the group, which is again we can find uh from the equation one because the tension of the roof is nothing but four times. So that's gonna be four times a day. And we already have the value of a as uh 1.96. 1.96 times four Gives us 7.84. Newton's as the value of the tension. Uh This is the value of the tension and then we have to find in parts C. There was the speed with which the hanging mass hits the floor if it starts from rest. And let's look at the 10 1 metre uh from the floor of the hanging basket. So in short, it has to travel one m. So, what's the speed after traveling? one m? That's what they're asking. So, we have a mask over here which is accelerating down At 1.96 m/s square. Ah And it hits the ground. So, it travels one m and we need to find out what is the corresponding speed when he does that when it does that? All right. Mhm. Uh What can we use here? Uh We don't know at the time. So, can we use the second equation of the motion because second equation of the motion is equal to we not d plus half a T square. So, from this equation we can at least get the value of T. So S. Is one uh definitely starts from rest, so uh this is zero and this is half times 1.96 times t square. So T Square is going to be to over 1.96 of the value of T. Com Sodas, Uh two divided by 1.96, Directed by 1.96. and square root of this is going to be 1.01, so 1.01 seconds is something which it takes. And we can also find that the final velocity we should be the north plus a tv in order is already zero, is 1.96 and T is one point The 01 city as one 01. So this is going to be equal to 1.91.01 times 1.96. That's going to be one 1.98 m/s. So this is the required value of the the lawsuit, even if it's the floor. Thank you.

So we can apply Newton's second law on the three masses. This would be the quick free body diagram of the system here, we can say that these some of forces in the Y direction for for mass someone would be equaling of course the first mass multiplied by the acceleration of the system. We could say this would be equal to M one M 01 G minus teeth of one Equalling then M sub one. Hey, we have the net force on the second block in the X direction. This would be equal to of course M sub two A. Substituting in we have teeth of one minus T. Sub two equaling M. Sub two. A. And finally the sum of forces on the third mass in the X direction equal and of course again, M C. Three A. We could then say that then TC two equaling M three a. So from this we can substitute one essentially into the other and say for part A. You can say teeth of one minus T. Step two. We know this is M. Sub two A. But we know that. Then we could substitute this would be equal to T. Someone -M. Sub three. Okay. and so solving for teeth of one piece of one would then be equaling two M. Sub three A. Plus um Sub to A. Taking this one. We can say that then M sub one G minus T. Sub one equals M. Sub one A. Or mm someone G minus I'm sub three A minus to A equaling um someone A. Or essentially solving for the acceleration. The acceleration would then be equal to M sub one G. divided by the sum of all three masses and be a little neater. Okay once we have that we can actually then Subsequute. We have then this would be equal 2.50 kg. Multiplied by 9.81 m/s squared are divided by then .50 plus 1.50 Plus 2.50 of course units of kilograms. And we find that then the acceleration a. of the system 1.09 meters per second squared for part B. We can find that tension. Say that um someone g minus tension, teeth of one equaling M someone A. Now that we know the acceleration And the tension sub one Would then be equal to um so one multiplied by g minus A. So we have then .50 kg Multiplied by 9.81 -1.09 units of m per second squared. and tease of one is found to be equal to four .36 Newtons. That is the end of the solution. Thank you for watching.

In this case, here is four k g. So the wet is for G This for this t one. So here we have the one we have got one kg so one G and the normal forces one g and we're gonna depend fiction later. The tension here is to you too. And then finally in the second object, Anjali, Streeto and Disease do g downwards so we can understand that when it accelerates, acceleration will be in the downward direction A in the upward direction A and on the left direction A and friction with be on the opposition def motion so that he disease fiction is mu times the normal force which is G So we have total resident force in the left direction is so we have for G minus T one equals for a We have G one minus mule G minus T do equals a or one a. And then for this object it is. Do you do my nest Ujiie? It was to a so we can sit the adult of them up. Can we have Dima? Yeah, we could have developed. You're meant to do with cancels out. So they stances out this cancels out on these at the Dunder did that is being added. So four g, the Studi so that 16 minus mu G six minus mu G. It was seven A. So you can find a quiz. 9.8 times mu is cedar 0.35 or six minus 0.35 years, five point 65 Do you want it by seven. So that acceleration comes out to be well going. 31 meter per second Squared now, once you have a from here, you confined to yuan equals 30 new done. And from here, you can find Kato equals 24.2. New done.

All right, let's solve problem 28 in Chapter five. So let's just will do the drawings one object at a time. So first we have this mass one, and it's on this table and we want the free by diagrams so we know it has gravity. And because it's on a table, it has to have some normal force that way. Doesn't just seek through the table. Uh, we know it's connected to a rope. It's being pulled by that rope with some tension and for friction, for table with friction, there would normally be a friction force here, right? But this is a very slippery freshness table, so this one inches Europe so you don't have to draw this, But keep in mind for later questions that might happen. Okay, Uh, and we'll just, you know, we know this this thing is clearly not falling through a table or somehow magically jumping up in the air. So this normal force just as equal fg Okay, so that's this problem so far. We know there from this problem that it's moving. It's accelerating towards it, right? So how much does it is? Its acceleration Well, f equals I made the only f we really care about this tension. And she called m one a one, right? Nothing terribly surprising there. Okay, here we have the weight connected at the bottom. This one has its own gravity force m two g, and it also has this tension, and so it's also accelerating. And so the summer forces here and we won't be consistent with our convention. So the if we're treating moving this way to the right as positive and were saying that this whole set up is something like, you know, saying connected here strong over here. And here's our object we should treat treat This direction is positive and which also treat this directions possible because they're sort of the deposit direction. It's just changed due to the police. So here we actually right and two gs positive, minus tension. And that's equal to I m too a to. And so we want to find, uh, what the tension is. And we want to find what the acceleration is. And so we have two equations, but we have thrown. In fact, we have tension exploration. Want explosion, too? Yes, that's three. And so the key is because they're connected by this single rope, right? They have to be moving together. Which means that a one has to be equally to that. Soldiers write us a okay. And so from this equation, we can write a That's just cheeky over m one. Okay, we can then play that over here. And so we get to G minus t equals no to on one turns t If we do a little bit of rearranging insult for tea, we get the tea is just and to G over one plus and to over one if we plug in all number. So this is into its nine and one. It's one just good old 9.1 should get 31.5 news. And then we know that the acceleration it's just t over one. Right. That's 31 point five Nunes over the five kilograms, and that's just 6.3 meters per second square. You've actually


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