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ssm Two objects $(45.0 \text { and } 21.0 \mathrm{kg} \text { ) are connected by a massless }$ string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find (a) the acceleration of the objects and (b) the tension in the string.

3.56 $\mathrm{m} / \mathrm{s}^{2}$

281 $\mathrm{N}$

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Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

for this question, we have to use a very uncommon trick to solve it. So instead of choosing AH reference frame like this, we will choose a curve with reference frame. By that time. In the following, these will be our ex access or y axis, as you like to call it. I call it the Y Axis. So everything that is pointing on this direction is pointing to the positive direction on everything that is pointing to the other. Direction is pointing to the negative direction. Then, by doing that, we can apply Newton's second law. So the wall system, by the old system I mean these together with this, then these results in the following. Within that force on that direction is the coast to the mass off the wall system. So M one plus m two times acceleration off the world system. Call it eight and now direct in that force, the net force is composed by four forces. The weight off the second part. Attention, attention again and they wait on the first part. And this goes as follows The weight number two points to the negative wider action. So my newsweek number two, the tension number two points to the positive Y direction, minus the tension on the one which points to the negative y direction. Note that both tensions are equal because the cable is considered here to be an ideal cable. So it can't be straight ID nor compressed. And then we have they wait force that were you one which points to the positive y direction in these easy of course to the war mass times excellently from offices them. They only have a cancellation here off the tensions and then we get though you want miners level, your true is equals to M one plus m two eight. Remember that the weight force is a given by the mass times acceleration off gravity, which is approximately 9.8. And then we got the following. The acceleration off the system is ECOWAS troops and one times G minus and choose times G, divided by m one plus m. Chu settled acceleration is he close to 21 times 9.8 minus 45 times 9.8 divided by 21 plus 45 and this resulting in acceleration off approximately minus 3.56 meters per second squared It means that the system we work salaried in these directions. So these part will go down in this part goes up and this happens with an acceleration off 3.56 meters per second squared. These is the absolute value off the acceleration. Now for the second night and we have to calculate one is the tension in that cable. For that, we apply Newton's second law either to this system or the system. So let me apply to the system doing that. I get the following Oh, and now I'm using these reference frame for the second item. Deasy's my Y axis. So it is worse for the first. And these is for the second item. Then on that why direction? The net force is equal to the mass am number one times its acceleration. The Net forms is composed by true forces Attention on the way to number one. Then we have the tension pointing to the positive direction miners, the weight number one pointed to the negative direction being equals to m one times eight. So detention is equals to m one times eight plus w 10 what is this acceleration off the weight number one and now note. That's the week. Number one. We'll go up according to our answer for the last item, then the acceleration a one is equals to 3.56 meters per second squared. So the tension force is Danny Coast to M 1 21 times three points, gift six plus and one, which is 21 times G 9.8, resulting in attention off approximately 200 and 81 new terms.

Brazilian Center for Research in Physics