3. A 3.50 kg block on a tabletop is attached by a string to...

3. A 3.50 kg block on a tabletop is attached by a string to a hanging block of mass 2.80 kg, as shown below. The Blocks are released from rest and allowed to move freely. I also apply a small opposing downwards force F = ¼ m1g on the block at an angle of ? = 30° above the tabletop but the mass m1 continues to accelerate towards the pulley. Find the Tension in the string and acceleration if the coefficient of kinetic friction, ?k, between the mass m1 and the tabletop is 0.45. [20 pts] Sketch and Label all forces. Define the coordinate axis for each mass.

Transcript

00:01 In squestion we are given that there's a 3 .5 kilogram block on the tabletop.

00:04 So this is 3 .5 kilogram and that is attached by a string to a hanging block which is 2 .8 kilograms as shown below.

00:15 So the blocks are released from rest and allowed to move freely.

00:17 I also by a small downward force which is half of one fourth of m1g.

00:24 So m1 is 3 .5 times g.

00:29 So this is the force.

00:31 Which is they apply in this direction and the angle is given as 30 degrees.

00:37 So this is 30 degrees.

00:39 So but m1 continues to accelerate downwards.

00:43 So let's say the acceleration downwards is a.

00:49 So we have to find the tension of the string because there is a tension over here.

00:53 There's a tension over here.

00:55 And the coefficient of kinetic friction is muke between m1 and the floor.

00:59 So there is definitely a friction over here.

01:02 As well.

01:03 Definitely there is an m2g over here.

01:07 This is the normal reaction and this is the m1g.

01:12 We've got to make the components of the force.

01:14 So this is going to be f cosine 30 and f sign of 30.

01:24 All right.

01:24 Let's see.

01:25 If we make the equations on this m1 for the vertical forces, is on m1 then i'm going to make a normal reaction must be equal to m1g plus f sign 30 f sign 30 degree and uh sign 30 is nothing but 1 over 2 and if is nothing but 1 over 4 m1g so this becomes a 9 over 8 m1g this is the value of the normal reaction it's not box it up for the timing this is the normal reaction.

02:07 Now why do we use that because we need to find friction force as well and that's where it will be used.

02:12 Now if we talk about horizontal forces, let's talk about m1 first because we are talking about m1.

02:21 So that's going to be t minus f cost 30 minus frictional force which is mu k times the normal reaction.

02:36 So over here we have mu k and the value of mucay is already given, that's 0 .45...

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