00:01
Hi there, so for this problem we need to determine the tension in the rope in terms of the weight in each case.
00:11
Now for case a we need to do first the diagram of forces, so we draw an axis.
00:27
This is the y -axis, this is the x -axis, and in the centers we draw the block.
00:39
And in here we will have, as you can see from the pipel.
00:42
Sure we will have two forces.
00:44
Detention upward and the weight.
00:53
Now this system is a dress.
00:57
It is not moving.
00:58
So when we apply newton's second law and assume all of the forces, we will have that the tension minus the weight is equal to zero.
01:08
So we can conclude in here that detention is equal to the block so this is a solution for this problem for par b we are given to blots as is shown in the figure b and they both has the same weight now with first i'm going to do the block at the left and later the block of the right now first as before we draw our aftsiz and for this first case, for the block at the left, we will have detention, that we call simply the tension, and the weight.
02:16
Now, we know that this system is not moving, since both masses have the same weight.
02:24
So, here, when we apply newton's second's law, we obtain sum of forces, give us the detention minus the weight is equal to zero.
02:33
So again from this, we obtain that the tension is equal to the weight.
02:46
Now for the mass at the right, as you can see, we are working with the same string...