Two blocks are connected by a "massless" string over a...

Two blocks are connected by a "massless" string over a "frictionless" pulley as shown in the diagram. (10 marks) 0.80 kg ?k = 0.14 60° ?k = 0.14 2.0 kg (a) Determine the acceleration of the blocks. (b) Calculate the tension in the string . (c) If the string broke, for what minimum value of the coefficient of static friction would the 2.0-kg block not begin to slide?

Transcript

00:01 In this question we have given two blocks and are both connected by a massless string over a frictionless pulling and in the first part of the problem we have to find the acceleration of these blocks so let the mass of this block that is m1 is equals to this is 0 .8 kilograms and m2 is 2 kilograms and we have given the value of muay that is equal to 0 .14 here and the angle that is theta is equal to 60 degree now i'm going to show the forces acting on this system.

00:34 So weight of this block is acting vertically downward that is equal to m2 into g here.

00:39 And if we draw the component of this weight, so in this direction this is equal to m2g, this is cost theta because this angle is theta here.

00:52 And in this direction this is m2g sine theta and the normal force in between the block and the surface is this is n2 here.

01:05 And there is a frictional force acting in this direction this is equal to let f k this is and this is a tension force in the system which is t here and on this block there is a frictional force acting opposite to this motion that is fk here and let the acceleration of the system is in this direction that is a so we can say that the frictional force is given by fk that is equals to mu k into this is let this friction force is equal to this is n1 and this is of f2 so if we calculate the friction force 1 so this is equal to mu k into m 1 into g so you can say it's equal to 0 .14 into 0 .8 into 9 .81 similarly second friction force that is fk 2 this is equal to mu k into n2 which is equal to m2g sine theta now we know that the acceleration of the system is given by that is equals to net force upon effective mass.

02:22 So we can see that the net force in downward direction is this is m2g cost theta minus the fictional force that is mu k into m2g this is sine theta minus mu k into this is m1 g upon this is m1 plus m2...

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