1. Three boxes of masses m, 2m, and 3m are connected as...

1. Three boxes of masses m, 2m, and 3m are connected as shown in FIGURE below with m = 1.0 kg. A force F?? = 12N is applied to accelerate the boxes toward the right. There is a coefficient of kinetic friction ?k = 0.1 between the boxes and the horizontal surface. Answer the following questions. a. (3 pts) Draw free body diagram for all three different masses showing and labeling all the forces present. b. (2 pts) Determine the acceleration of the boxes c. (5 pts) Determine the tensions T1, T2 and T3 in the ropes m Box 1 2m Box 2 3m Box 3 T1 T2 T3 F?? = 12N

Transcript

00:01 Hello everyone, we are going to understand this question here given in the question given 3 boxes of mass m 2m and 3m masses of the box m 1 is equal to m m 2m and 2m and m 3m which are connected as shown in the figure here m is equal to 1 kg a force f is equal to 2m and 4 f f is equal to 2 is applied to accelerate the box towards the right.

00:48 There is a coefficient of kinetic friction between the box and horizontal surface.

00:54 Here, mu k is equal to 0 .1.

00:58 So we have to find, we have to draw the free body diagram, we have to determine the acceleration and tension in the strings.

01:06 So here, for box 1, weight of the box, that is m1 into g, t1, t1, and here m1 into g will be balanced by the normal reaction, that is n1.

01:21 And there is a friction force as here block is moving in rightward direction with acceleration a, so here friction force will be in left side, that is f1.

01:33 Similarly for box 2, weight of box 2, that is m2 into g, and normal reaction which is in upward direction, tension in the string that is in rightward direction and tension t1 which is in leftward direction on the box 2 and there will be a friction force that is f2.

01:57 Similarly for block 3, weight of the block 3 that is m3 into g which is in downward direction, reaction force which is n3 which is in upward direction, tension t3 which is in rightward direction and t2 is in leftward direction.

02:13 Friction force is in left direction.

02:17 That is f2.

02:19 Now it is moving with acceleration a.

02:23 So let the net acceleration be a.

02:34 The net acceleration be a.

02:37 Now first of all calculating friction force f1 is equal to mu is 0 .1.

02:47 V1 is equal to mu is 0 .1.

02:49 Value of n1 is is m1 into g now substituting the value in this so 0 .1 into value of m1 is m so 1 into 1 into 9 .8 after calculation we will get 0 .98 newton this is the friction force f1 now calculating f2 which is mu into n2 that is equal to mu into m2 into g.

03:22 Here m2 is 2m.

03:26 Now substituting the value in this 2 into 0 .1 into 1 into value of g that is 9 .8.

03:37 After doing calculation we will get f2 is equal to f2 is equal to 19 .6 and here decimal will be after 1 after 2 number.

03:59 This is the value of f2.

04:03 Now calculating f3.

04:07 Fxion force on the third block that is f3.

04:09 So mu into n3.

04:11 Value of mu is that is 0 .1.

04:13 Value of n3 is m3 into g...

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