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15/2015) As shown in the figure weightless rope is wound on identical cylindrical pulleys of mass m = 6 (kg) What is the tensile force (N) on the rope when the sysl...

Question

15/2015) As shown in the figure weightless rope is wound on identical cylindrical pulleys of mass m = 6 (kg) What is the tensile force (N) on the rope when the syslem is released: The shalt ol the pulley above is frictionless: g = 10 (m /s7)4)6B) 8C)0D2E) 16

15/20 15) As shown in the figure weightless rope is wound on identical cylindrical pulleys of mass m = 6 (kg) What is the tensile force (N) on the rope when the syslem is released: The shalt ol the pulley above is frictionless: g = 10 (m /s7) 4)6 B) 8 C)0 D2 E) 16



Answers

A 1.5 kg block is connected by a rope across a 50 -cm-diameter, 2.0 kg pulley, as shown in Figure $\mathrm{P} 7.76 .$ There is no friction in the axle, but there is friction between the rope and the pulley; the rope doesn't slip. The weight is accelerating upward at $1.2 \mathrm{m} / \mathrm{s}^{2}$ What is the tension in the rope on the right side of the pulley?

I had to see the diagram you to consider for the questions. So that is it block having mass is pulled by a rope of mass smaller, right small with the force of Okay, what we can do over here in the first place, I need to say that it is saying. So basically we can clearly see that the small sort of the block robbed by some masses. So it will most people where the gravitational force so we can believe we will be reached at the point when it had obviously don't have 10 20 they have uh you know the side wise point of the block any, the point mass is pulled upwards so that MG balances out uh group attention. So and uh it is something that Tencent have a component in that direction. So it is broke. This rope is sagging. Okay sense. Okay. And find the explanation of the rope and block planned execution of rope and the block. Okay, so uh So the explanation the only 4:15 of the original component. So basically a solution, what if we're taking both of the system? That's a lesson with a mask times of excellence and will be network for the radio six innocent. It will be a phone M plus and right. And the next part calculate is that the force on the block from the rope? Mhm. Force on the block from the very simple, very very simple question. What we can do over here is that course on the block due to road that will be nothing but a mass stamp of excellence and were responsible for the motion of the block. Okay, so this will be my last effort to listen to mass types of excellence and explicitly department F upon employees and and plus a small handful M. F upon and plus them. So this will be the fourth in the next part, the uh the tension in the rope at its midpoint. So 10cent in the room, a robe at the midpoint will be nothing. But at the middle point at the middle point, this Jensen should be responsible for the rest of the motion of the mask. To strengthen T equals nothing but mass. M plus half of the rock mass. Because this is the middle point into a so it will be M plus a small um upon to into F upon and bless them. Okay, thank you.

This problem covers the concept of Newton's second law of motion and for the heavier objects, which is moving Don Wood with the initial velocity or the initial speed zero, we can right. The vertical distance travel is, if you will, into half of the leader acceleration A into T square. And from this equation we can calculate the linear acceleration is two times y upon t square. No substitute the value. To calculate the acceleration so equals two and two. Why is 75 centimeter or 0.75 m upon five seconds square, the acceleration is zero point 06 m or second square. So the heavier mass is moving downwards with acceleration A and the farmers will be moving upward with the same acceleration. No part. To solve this problem with straw, the freebird diagram after heavier mass. So this is the mass capital M. The forces acting on the masses. The weight, which is acting word for downward, that is MG. And the tension even which is acting at the airport. So from the Newton's second law of motion, the acceleration of the heavier masses along the downward direction. So we can write the net force along the right direction, along the direction of motion. That is M. G. Minister attention to and equals M. A. Are that ancient even equals am into G minus A. Now, let's substitute the value to calculate the tension, so even equals the mosque. M s 0.5 kg in two G. S 9.8 m four seconds square. And the acceleration is 0.6 Need us for a second square. So the tension to evenness, 4.87 noted. Now part C. To solve this part, we need to draw the free world diagram of thomas swollen, the forces acting on the mass, small and mr ancient T two, which is a strong word, career port, and the weight which is acting vertically Rhonda. And the acceleration of the object is a longer vertically upward direction. So from the 12th law of motion we can write data minus MG equals m a r t two equals m n two G plus a. I've showed the value so the tension T two equals the mass small um is 0.46 kg into 9.8 plus 0.6 meters for second square. So the tension T two equals 4.54 No, no party for no sleeping condition. The angular acceleration of the bully is the financial acceleration upon the radius of the police. Now substitute the value to find the other acceleration. The tangential acceleration is 0.6 Need us for second spare and the radius is five centimeter. Or we can write 0.5 meat. So the angular acceleration of the police 1.2 radiance four seconds square no card E from the Northern Second law of motion. The net talk on the police is the rotational inertia. And it's uh angular acceleration. And the net talk is G two into our minus Stephen into our and that equals I alpha. So from this we can write our equals two to minus Stephen into our upon alpha. So the rotational inertia of the police Tito is four point 87 million minus Stephen is 4.54 newton into the radius 0.5 Media upon them with acceleration is 1.2 radiance for second square. So the rotational inertia of the police, one point create into And this minus two kg middle square

In this problem we have to determine the exploration of this system. So Newton's second law is and the net force is mass into the excavation. So we have from the figure the net force acting on the mass. M one is T one minus M one G. This is a question one. And the net force yeah acting on mass. Everyone is everyone. Mm. This is the equation to So using both the equations we can say Using Creation one and two We can say and one a. is equal to 51- and one G. So from this T one is equal to M. One A plus M one gee. This is a question three. And then that force acting on the mosque two is I am to G -62. This is a question for and the net force acting on them as two is um to a Now using equation Using equation four and 5 we have em too is a call to M two G minus T two. So T two is M two G-. And to desist equation six. And the equation for net torque on the police half mm are square multiplied by a upon. Hi this is equation seven. And the equation for net torque is this is the equation for net torque on literally. Yeah. And the expression for net torque is um RT to minus RT one minus style. This is equation eight. So using equation seven and eight. Mhm. Uh huh. Rt to minus R. T. One minus so is equal to have an eye square multiplied by upon our. So from here the two minus the one minus how are is equal to half M. This is equation nine. So now you think equation three, six and nine we have M two G minus and one g minus tao upon us, he recalled two and one plus M two plus half. I am might be fired by a So from here is a Dorito. I am to G- and one key minus towel upon our of born and one I am too. Let's have Yeah, so We have M to 8.8 Kg. She is 9.8 metaphor 2nd sq my nurse and one is 10.4 kg multiplied by 9.8 metre for second square. My style is point 35 new 10 m upon. It is .15 upon and one is 10.8 K drip plus M two is point for Fiji. Let's have M. S 0.2 G. So from here is 1.2 m for a second square. Therefore the isolation of the system is 1.2 metaphor second square.

We have a there's a box resting on the table and then another box hanging from a string over a polar here. So I've drawn free by diagrams of these, Technically this one is kind of down here. Um So this is the box that is free to move horizontally. This is the one that's free to move vertically. And here's our polling. So we have to wait and attention and napoleon. That tension is also in this side of the of the rope that's going around the polio because the pulley has inertia. Uh this tension is not necessarily the same as this tension and but this tension is the same as this tension because that's in the same part of the rope and then that tension is applied to this and so positive. I'm going to say to the right for this block down for this one and counter clockwise for the polling. Have a friction for us acting on here. And then these are reaction force is at the center of the polling. We're told that this one block weighs 12 kg. This one the one hanging weighs five. The poli ways to and the police has a radius of 0.25 m. And I guess they tell us that it's a frictionless surface. So friction forces zero. So we don't have to worry about that. So looking at this body here, we have and if this is positive, we have minus T. One plus W. T. Equals M. Two A. And we're told that this um the belt around here or there rope is um not slipping on this pulley. So we know that the angular acceleration of the poli equals the acceleration of this block divided by our. And we also know that that acceleration of this block must be the vertical acceleration of this block must be the horizontal acceleration of that block. Let's see here, looking at this block in the horizontal direction, we just have T. Two. He calls M one A. Because The Friction Forces zero. The pulley, the is assumed to be a uniform cylinder. So its moment of inertia is one half M two R squared and a moment balance on this. About the center here is T one R minus T. Two R equals J. The mass, moment times the angular acceleration. No, we can do a bunch of basically, we need to eliminate things. We can solve plug, take this and plug it in here. Take this self 31 plug it in here. So for T to plug it in here, you know, make the weight M two G. Use this relationship here and here and we get an equation that we can then solve for A. And I didn't go through all the algebra here, but in the end A comes out to be M. Two G. All over the quantity one half M. Three plus M. Two plus M. One. So we can see here that um uh let's see here, Say M. one and M. two cm one is 0 And I'm three or 0. So this is zero and this is zero. Then this cancel, this goes away. And then a. Is just because this is just free falling. So that makes sense. If M. Two or zero then A. Is zero. Because if there's no wait down here then there's nothing to pull getting this um to start moving anyway and we can see that you know as as um you know, M wanted them to increase. Then it goes down. So in this case when we plug in our values we get 2.72 m/s squared. Once we have that we can plug back into here Knowing the weight we get that detention. And this part of the vertical part of the cable, there's 32.6 newtons. And then we can plug this into here and we get the tension on the horizontal part of the cable is 35.4 newton's. So you can see that they are different. Different because um if this was zero, so j if this is massless And this was zero, then they have to be equal. But because we have mass in the pulley and it's accelerating. There's a net torque on the polling. Now they also asked us for the reaction forces in the poli. And so if we just do force balance in the police center of mass is accelerating. So our acts must equality too. And ry must equal T. One plus the weight of the poli. And that's 55 new


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