00:01
In this question, we are given that a block mass mb is on a horizontal surface with a coefficient of kinetic friction mukeye, which is 0 .54.
00:14
And there is a massless frictionless pulley about which a string gets connected to a mass ma.
00:21
And this mass ma is on a frictionless inclined plane.
00:26
And this inclined plane is at an angle 32 degrees.
00:29
So this is the situation shown in this diagram.
00:33
So with this arrangement, we are asked to find what is the tension in the string as well as the acceleration of the blocks.
00:40
So since we see that the mass ma is greater than mass mb, and that's the one that's going to drive the motion.
00:47
So let's assume that both the blocks are moving in this direction.
00:53
And both accelerations will be same because they are connected to the same string.
00:57
And there is no slippage, there is no slack in the strings.
01:01
So the displacement velocity and acceleration of both these blocks will be same in the given direction.
01:08
And so the accelerations of both blocks will be same.
01:11
And similarly, because the pulley is mass less and frictionless, there is no torque or inertia of the, there's no torque that the pulley can maintain or talk differential because it has no inertia, which means to say that the tension in the strings will be same.
01:26
Let's call it t on both sides so with this let's draw the free body diagrams for both these blocks so for the b block there's a mb times g weight force there's a normal force due to that because there's horizontal plane which creates a reaction force and that will create a frictional force mu kfn which will be against the motion of the block since we are assuming the block is moving towards the left there will be a tension force and since this is moving at an acceleration of a there will be a pseudo force mb a which will help us create this into a inertial frame of reference so that is the free body diagram for an accelerating body mb and then we also draw it for let's say the body m a so it has a weight ma g there is a cost theta component there is a so this will be theta and if you want to to see how this is theta and draw two lines here and draw this line here this is 90 degrees and this is 90 degrees so if this is theta this is 90 minus theta and this is 90 minus theta so that's how you figure out this angle to be theta so i'll just remove this back so this uh back coming back to the free -parated diagram this ma g sine theta component is this one there will be a normal force which will be fn prime which we don't need because we don't need because there's no friction over here.
03:13
There is a tangent force t in this direction and since this also is accelerating down the inclined plane with an acceleration a, there will be a pseudo force m .a times a in the opposite direction.
03:26
So these are the different forces on both these bodies.
03:31
So let us start making equations out of these.
03:35
So let's first do block b.
03:38
So block b.
03:42
So fn is equal to m b g which is very straightforward this force is balancing out the weight force so there's nothing the other force in this equation and the horizontal equation gives us that tension t is equal to m b a which is a pseudo force plus mu k times fn and fn we can get from here straight is m b g so we know this these terms let's simplify it and write it in terms of that mb is 2 .2 a plus mu k is 0 .54 multiplied by 2 .2 multiplied by 9 .81 which is acceleration to do gravity...