Question 2 three connected masses are arranged with two...

Name: question 2 three connected masses are arranged with two pulleys of negligible friction on a table as shown in the diagram below determine the tension in the left cord f11 and the right cord 67227
Uploaded: 2025-01-10T13:37:20-08:00
Duration: 12 min 22 s
Channel: Vishal Gupta
Description: question 2 three connected masses are arranged with two pulleys of negligible friction on a table as shown in the diagram below determine the tension in the left cord f11 and the right cord 67227

QUESTION 2 Three connected masses are arranged with two pulleys of negligible friction on a table as shown in the diagram below. Determine the tension in the left cord F_(11) and the right cord F_(12) after the system is released and begins to accelerate.

Transcript

00:01 Here in this given problem, first of all we redraw the diagram.

00:07 This is the fixed support at which the largest pulley is hanging, which is holding other two pulleys at the two ends of the string passing over it.

00:31 Now these pulleys, this is pulley a, this is pulley b, these are further holding two blocks each suspended.

00:42 At the end points of the string over the police.

00:46 This is 1 m .kg.

00:48 This is 2 means 2kg.

00:50 This is marked as 3 means 3kg.

00:55 And finally, this is marked as 4.

00:59 So its weight will be 4kg.

01:00 So first of all, we draw their weights.

01:03 Starting with this, that is 4g acting vertically down, 3g acting vertically down, 2g acting vertically down, and 1g means g also acting vertically down.

01:20 Tensions in the strings, starting with this t1 in both of the segments here.

01:33 T2 in the uppermost pulley in the wire passing over uppermost pulley t2.

01:39 And then this is t3 here.

01:45 Okay.

01:46 Suppose block 4 is falling down with an acceleration.

01:54 We begin with the largest pulley.

02:00 And we assume this pulley that is falling down with an acceleration ab.

02:06 So pulley a will be moving up with the same acceleration a.

02:11 Then we will get the accelerations in 4kg system that is a4.

02:22 But we will, we cannot give them the final name a4.

02:27 So first of all, first we find a and ab, for which using free body diagram of the pulley b, which will be falling down due to the weights suspended on it, 3kg and 4kg.

02:52 So 3kg plus 4kg means 3 plus 4g, that should be equal to tension in the string.

02:59 That should be more than tension t2 in the string.

03:02 So 3 plus 4g minus 2 that should be equal to 3 plus 4a that is a b.

03:11 Or we can say 7g minus t2 that is equal to 7a.

03:17 A b and a, a both will be equal as the string is same.

03:24 So we use only a for that.

03:26 That is equation number 1.

03:31 Now using free body diagram of pulley, a which will be moving up with the same acceleration.

03:41 So for it, tension t2 will be more than the weights suspended on it, 1 kg plus 2kg.

03:48 And should be equal to 3 times of a.

03:54 Means t2 minus 3g is equal to 3a.

03:58 That is equation number 2.