88. Two blocks are sliding to the right across a horizontal...

88. Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is 3.0 kg. In Case B the mass of block 1 (the block behind) is 6.0 kg, and the mass of block 2 is 3.0 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 5.8 N does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks. Block 1 Block 2 m1 = 3.0 kg m2 = 3.0 kg Case A Block 1 Block 2 m1 = 6.0 kg m2 = 3.0 kg Case B

Transcript

00:02 Hi, here in this given problem, there are two blocks kept over a surface in two cases.

00:10 There are two cases.

00:13 In the first case, the blocks are identical in contact with each other.

00:23 In the second case, the blocks are not identical.

00:33 This is block 1, this is block 2, here also this is block 1, this is block 2.

00:40 Its mass that is m1, its mass that is m2, here also m1, m2 this is case a, this is case b.

00:51 No friction for the first block in both the cases but friction, force of friction, fk for the second block in both the cases.

01:03 Fk in the first case the masses are m1 is equal to m2 is equal to 3 .0 kilogram in the second case m1 is 6 .0 kilogram and m2 is 3 .0 kilogram in the first part of the problem we have to find force contact force which is being a applied by the two blocks on each other, that force will be equal and opposite as per newton's third law of motion.

01:43 So if this is f at 2 due to 1, here it will be f at 1 due to 2.

01:49 Here also this is f at 2 due to 1 and here it will be f at 1 due to 2 magnitude equal but directions opposite.

02:02 So in the first part of the problem.

02:04 We have to find the values of f1 or f2 or f2 whatever we can say so using first of all i'm doing it for case a using free body diagram of block 1 there is only one force acting on it this is f1 2 and acceleration that is towards right in both the cases so using free body diagram of block 1 the force f1 .2 is equal to using newton second off motion, net force is equal to m into a.

02:55 So here it is m1 into a.

02:59 And using free body diagram, here this equation will give me the expression for acceleration, a is equal to f at 1 due to 2 divided by m1.

03:16 And using free body diagram of block 2, at block 2 there are two forces f21 and fk.

03:31 So f21 minus fk, that should be equal to m2 into a or we can say f21 minus fk is equal to m2 into a or we can say f21 minus fk is equal to m2.

03:50 M2 for acceleration that is f1 2 by m1 or f1 2 may be written as minus f2 1 so this is m2 by m1 and here it is minus f2 1 as already we know these forces are equal in magnitude but opposite in direction so finally this is f2 1 plus m2 by m1 into to f21 is equal to fk and we know the value of fk that is given to us as 5 .8 newton for both the cases...

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