Consider two blocks M1 and M2. M1 is placed on a table with...

Consider two blocks $M_1$ and $M_2$. $M_1$ is placed on a table with friction and attached by a string to $M_2$. The string passes over a massless, frictionless pulley so that $M_2$ hangs down, as shown in the figure. $M_1 = 18$ kg and $M_2$ is unknown. The coefficients of friction between $M_1$ and the surface are: • static friction: $\mu_s = 0.36$ • kinetic friction: $\mu_k = 0.12$ When the blocks are released, the system is observed to remain at rest. What is the largest possible value for $M_2$ such that the system remains at rest?

Transcript

00:01 Hello everyone, we are going to understand this question here from the given data in the question.

00:11 From the given data in question, m1 is equal to 0 .45 kg, m2 is equal to 0 .82 kg.

00:44 Coffreciant of static friction between the blocks.

00:56 Coffreciant of static friction between blocks that is represented by m2.

01:19 Mu into s, that is 0 .56.

01:24 And coefficient of kinetic friction, coefficient of kinetic friction, that is represented by mu k, which is 0 .34.

01:50 Now, m1 will not slip if, now m1 will not sleep if fs is greater than equal to m1 into a.

02:16 M1 will not slip on the block m2.

02:20 So we can write mu s into m1 into g greater than equal to m1 into a.

02:40 Or we can write mu s into g must be greater than equal to a, then it will not slip.

02:51 So we can write value of mu s is 5 point, sorry, 0 .56.

03:07 0 .56 into 9 .8 that is greater than a.

03:21 Then block m1 will not slip on block m2.

03:25 So after calculation we will get a must be less than equal to 5 .488.

03:36 Now firstly let's draw the fvd.

03:45 This is m1 and this is m2.

03:52 When this block will move in the right word then there will be a friction force which is on the left word that is represented by f s1 and there will be another friction force that is fs2 and it is moving with acceleration a and friction on the upper block that is fs 1 with because of this friction force this block is moving with acceleration a so there will be a pseudo force in the left direction that is m1 into a.

04:29 So block m1 will not move if m1 mu if friction force fs1 is greater than equal to m1 in 2a.

04:42 In this condition it will not move.

04:45 So from this we have calculated a must be less than 5 .488 meter per second square.

04:53 Now for second block we can write here tension in the strength this is tensor in the strength t minus t minus f s 2 t minus f s 2 is equal to m1 plus m2 into a now substituting the value in this so we can write t is equal to m1 plus m2 into a plus mu k into a plus mu k into m1 plus m2 into g...

Question

Please give Ace some feedback