00:01
Hi, as given in the figure we have three boats of masses m1, m2 and em3 that are connected by two ropes here and has been pulled by a third rope with an external force.
00:12
Which is to find the tension in all the three ropes.
00:16
Let us mark the tension in this rope as t1.
00:21
Here the tension can be taken as t2 and here could be taken as t3.
00:28
If all the boats gets pulled in the same direction, as the applied force with an acceleration as a, then applying newton's second law of motion, f is equal to m .a.
00:41
We can now write the external force applied to be equal to the mass, which is the sum of all the masses of boards, m1 plus m2 plus m3, into same acceleration a.
00:56
Now for mass m1, for first boat, the tension t1 acting on that boat, can be written as m1 into a.
01:09
Only one rope is connected to that board hence tension will be equal to the force external force m1a on that object.
01:20
For mouse m2 since it has been connected by two ropes the tension t1 acting on this boat m2 will be towards left side.
01:33
So we can write t2 minus t1 is equal to m2 into a.
01:40
T2 which is acting towards right and t1 acts towards lift...