In the drawing the rope and the pulleys are massless and...

In the drawing, the rope and the pulleys are massless, and there is no friction. Find $(\text { a) the tension in the rope and }(b) \text { the acceleration }$ of the $10.0-\mathrm{kg}$ block. (Hint: The larger mass moves twice as far as the smaller mass.)

Transcript

00:01 Hello, we have a question in which it is the system set up which is being given that a 3kg mass is suspending with the help of the string and this is a table on which 10 kg mass is placed and all are joined by a string and pulleys so all the strings and pulleys are massless so we need to find certain things let us suppose this to be mass m1 which is given as 3 kg m2 which is 10 kg now since all the strings are the same so masses should be equal to so the tensions in each the string will be equal so t this is also t this is also t now one more thing is being given that the that's when smaller block smaller mass moves a distance x so this larger block will move to a distance 2x okay okay which means if the displacement of this larger mass is 2x smaller mass is x so acceleration will be directly proportional to the displacement so if this is acceleration a this has to be acceleration 2a way so now if you draw free body diagrams let us suppose for smaller mass 3 kg so this will be w and there are only two tensions and for this larger mass that is 10 kg mass this is tension and this is the weight and this is the normal reaction by the table okay so and here this is 2a acceleration and this is excretion a so let us write the equations of motion since the acceleration is towards downward so net force must be towards downward so w minus t plus t will be equal to mass m1 into a so w minus 2t will be equal to m1a this is equation number one so for equation number two there is a no movement in vertical direction for this and horizontal direction it is moving towards the right word so there is only one force which moves it towards right word that is a tension so t will be equal to simply t will be equal to 2m a and let us write m2 so let us plug in the value of this what we say acceleration from here let us suppose that we have to solve problem number a in which we need to find the tension in the rope so for to get the value of tension we should get the value of exhalation from here and then plug in back to equation number one so exclamation will be t by 2m2 let us plug in equation number 1 w minus 2 t is m1 into t by 2m 2 m2 okay so it will be w equal to 2 t plus m1 t by 2m2.

03:56 Let us take tension t, 2 plus m1 by 2m2...

Question

In the drawing, the rope and the pulleys are massless, and there is no friction. Find $(\text { a) the tension in the rope and }(b) \text { the acceleration }$ of the $10.0-\mathrm{kg}$ block. (Hint: The larger mass moves twice as far as the smaller mass.)

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