a-5-kg-box-is-connected-to-a-pulley-with-rope-in-the-diagram-shown-below-if-the-radius-of-the-pulley

Question

A 5 kg box is connected to a pulley with rope in the diagram shown below. If the radius of the pulley is 0.5 m, what is the torque generated by the box on the pulley?
(A) 10 $\mathrm{N} \cdot \mathrm{m}$
(B) 25 $\mathrm{N} \cdot \mathrm{m}$
(C) 50 $\mathrm{N} \cdot \mathrm{m}$
(D) 75 $\mathrm{N} \cdot \mathrm{m}$

Sachin Rao · Accepted Answer

so long we have given a five kg box is connected to a fully support. This is a police and this is a box of Mars five kg. It is connected with a rope. Two police and we have given the areas of police 0.5 m, 0.5 m. We have to find what is the torque generated by the box on the police. So we know that target is equal to R cross F. We have f is the force. So force is equal to mg and this vertical stands Should we call to areas so we can find a torque will be equal to There is 0.5 into fourth mg just five and two 10. So this will be equal to 25 Newton meter. This is the magnitude of torque. So now if we see the options option, we is correct. Option B is correct. I hope I answered your questions. Thank you.

Salamat Ali · Answer

the door Police by master taste to a pulley. Improved unease Down is a call to our times f It is the rate of the box. So that is a cool two mg rmg substituting the values we have all reason 0.5 masses of five kg times the traditional extradition approximately 10. This gives us the dork off 25 new two meter. You didn&#x27;t meet her?

Eric Mockensturm · Answer

we have a there&#x27;s a box resting on the table and then another box hanging from a string over a polar here. So I&#x27;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&#x27;s free to move vertically. And here&#x27;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&#x27;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&#x27;s in the same part of the rope and then that tension is applied to this and so positive. I&#x27;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&#x27;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&#x27;s a frictionless surface. So friction forces zero. So we don&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s see here, Say M. one and M. two cm one is 0 And I&#x27;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&#x27;s no wait down here then there&#x27;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&#x27;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&#x27;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&#x27;s accelerating. There&#x27;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&#x27;s 55 new

Averell Hause · Answer

would be the free body diagram for the system. We have a mast up. One would be the 3.0 killer grandmaster, and massive two would be the 1.0 kilogram mass. We know that here for part a way, we&#x27;re going to apply noon, second long to the x axis of each box So we can say am sub to G Ah, sign of Fada minus t would be equal to M sub two times A. I&#x27;m so rather weakens to simply say F plus t would eat Will m sub one a. Again, these two are connected. Therefore, they&#x27;re gonna have the exact same acceleration. We&#x27;re gonna add these two together and then solve for the acceleration of the system so we can say m sub too. Rather, the acceleration will equal and sub to G sign of data and then this would be plus F. This would be divided by m sub one plus amps up to and weaken solve. So they would be equal to one 0.0 kilograms times 9.80 meters per second squared multiplied by sine of 30 degrees and then plus 2.3 Newtons bided by the total mass of one plus three. So four points or kilograms, and we find that the acceleration here is 1.8 approximately 1.8 meters per second squared. We&#x27;re going to a plug back into the first equation. Just simply could it&#x27;s easier. And we can say that here, the attention force would then be equal to m someone, eh? Minus the force f. And this would be equal to 3.0 times 1.8 minus again. Ah, here it would be minus 2.3. Let&#x27;s get rid of this. You don&#x27;t need that parentheses there on then here we find that attention Force is going to be equal to approximately 3.1 Newtons. This would be our attention forced. This would be an acceleration of the system. So that would be those would be your two answers for part A. Now, for part B, we want to find the essentially the critical case where F has reached the max value, causing the tension here too, then vanish. So, essentially the first equation the first part of the equation, Um, in part A. We can say that then the acceleration of equal G sign of 30 degrees, and this would of course equal 4.9 meters per second squared, or simply half of 9.8. This implies that, of course, T equals zero and then that the force would simply be equal to massive one, eh? Given that the tension force has disappeared. So this would be 3.0 kilograms multiplied by 4.9 years per second squared and this is equaling 14.7 Newton&#x27;s. This would be essentially the we can say critical F ah, for the tension to disappear. That is the end of the solution. Thank you for watching.

Meghan Miholics · Answer

in this question, we are told a horizontal rope is tied to a 50 killer and box and the box is on frictionless ice and were asked to covet the tension in the rope in a variety of cases. Now, before we start to calculate the tension, the force in the rope. Ah, what I would like to do is do a free body diagram of this object. Now, since we&#x27;re really only interested in the motion in the X direction, I&#x27;m gonna leave off the forces in the Y direction. So think about the forces in the extraction here. First of all, the Isis frictionless. So that means that there is no horizontal friction. There is no friction force acting on this object. And we know that there is a tension for us pulling on the objects horizontally. So the changing horses, actually the Onley Ford&#x27;s in the X direction that&#x27;s acting on this box. So any force that acting on this object in the extraction is completely coming from that tension force. All right, so let&#x27;s look at our different cases here. So the first case is what is the tension force when the box is at rest so at rest means that the velocity is zero. But it also means that the acceleration is zero right? The velocity is zero and it&#x27;s not changing. So this will imply that the force is also zero right? So you can think about it using Newton&#x27;s first law, our second law. When the acceleration is equal to zero, the force acting on the object must also be equal to zero. Okay, so what does that imply? It implies that the tension in the rope is also equal to zero. Okay, Um, there is no other fourth acting in the X direction. So if the total force is zero, then the tension also has to be zero. If there was friction force than it would be possible. You know the friction forces there balancing out attention, forced attention Force could be non zero. But in this case, since there is no friction force, if the total forces zero, the intention has to be zero as well. Okay, second case the box lose at a city five meters per second. So our speed or velocity here is five meters per second. And it says that the boxes moving at a steady rate So that means that the acceleration is zero for this case as well. The box is moving at a constant speed, so acceleration is zero since acceleration and zero. That means that the force total force acting on the object is also zero. And since tension is the only force acting in the X direction, tension must be zero in this case as well. So it&#x27;s a little bit odd something to think about, you know, moving a having an object, move without any force at all. Ah, but remember, this is frictionless ice. So there&#x27;s nothing pulling back on the box or something. They&#x27;re slowing it down. So there doesn&#x27;t need to be a force there, keeping it going per se and then finally in part C, we&#x27;re told the box has a velocity or a speed of five meters per second and that the acceleration is five meters per second squared. Okay, so now we actually have a little bit of something to calculate here. So if the acceleration is five meters per second squared, we can calculate the force. We were told that the mass of the object is 50 kilograms. Sorry. I should really write down what? The formula is amusing. This is F equals M E right? Newton&#x27;s second law. So the force acting on the object in this case is 250 mutants. The tension is the only fourth acting in the X direction. So tension itself is that 250 Newtons it is providing that force that is moving the box. So this is our final answer for part C and Part B and part A were both zero.

Averell Hause · Answer

so we should first find the acceleration of the entire system. So we can say that the total mass m of the system is gonna be 80.0 kilograms. And we can say that, um, the net force on the system is the force of gravity acting on the total overhanging mass. So we can say that the overhanging mass massive BC would be equal to 50.0 kilograms. And so we can say that the acceleration of the system is going to be equal to the mass. The total overhanging mass times G divided by capital m the weight of the system. This is gonna be equal to essentially 5/8 of 9.8 meters per second squared. And this is equaling 6.1 25 meters per second squared. Now, given this, we can say that here, we&#x27;re going to apply Newton&#x27;s second law to block C. And then we can say that the mass sub cg minus the tension of the total overhanging mass we can say this would be equal to mass sub c A. And so when we&#x27;re trying to find t&#x27;s A B c ah, we can say that here the tension in the cord connecting a rather BNC we can say is going to be equal to it would be m subsea, multiplied by G minus a substituting in. We find that here this is gonna equal 36.8 Newtons four party. This would be your answer for Part B. However, we&#x27;re going to then use Equation 2 15 and we know that Delta X plus rather Death X equals the ex initial t plus 1/2 hey t squared and we can eliminate that term and we find that the total X displacement would be 0.191 meters. This would be our final answer for part B. That is the end of the solution. Thank you for watching.

Problem

Which of the following objects has the greatest r…

Problem 32 Easy Difficulty

Answer

Related Courses

Cracking The AP Physics 1 Exam

Related Topics

Discussion

Top Educators

Recommended Videos

Watch More Solved Questions in Chapter 14

Video Transcript

Robert Franek

Cracking The AP Physics 1 Exam

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Cracking The AP Physics 1 Exam

Robert FranekCracking The AP Physics 1 Exam

Robert Franek

Cracking The AP Physics 1 Exam