A conveyor belt moving slowly at a constant speed passes...

A conveyor belt, moving slowly at a constant speed, passes through a small window in a wall which is high enough to accommodate a single block, but a careless employee stacks two blocks on the belt. The one at the bottom, block 1 which has mass m1=m1=3.43kgkg, continues, without slipping against the belt, through the window. (The belt continues without slowing.) The one on the the top, block 2 with mass m2=m2=1.97kgkg, presses firmly against the wall and slips across the lower block. The same coefficients of friction, μs=μs=0.55 and μk=μk=0.21, apply between the pair of blocks and between the block and the belt. Weights, if required, will be denoted with the corresponding subscript of the block as F⃗ g,iF→g,i, for i=1,2i=1,2. The normal force exerted by object aa on object bb, if required, will be denoted as F⃗ n,a→bF→n,a→b for a,b∈{1,2,belt,wall}a,b∈{1,2,belt,wall}, but a≠ba≠b. The force of kinetic friction exerted by object aa on object bb, if required, will be denoted as F⃗ k,a→bF→k,a→b for a,b∈{1,2,belt,wall}a,b∈{1,2,belt,wall}, but a≠ba≠b. The force of static friction exerted by object aa on object bb, if required, will be denoted as F⃗ s,a→bF→s,a→b for a,b∈{1,2,belt,wall}a,b∈{1,2,belt,wall}, but a≠ba≠b. What is the magnitude, in newtons, of the force of friction exerted by the conveyor belt on the block on the bottom, block 1?

Transcript

00:01 Hello students, in this question there is a conveyor belt moving, there is a box here and there is another box here on top and there is a thread connecting this.

00:13 So because of this thread and because of friction between these two surfaces, this box is not moving but the conveyor belt is moving.

00:23 So there is a friction happening between.

00:27 So now we can say that this force is being balanced, the force by the movement of conveyor belt is being balanced by the force in the tension in the string and the frictional force in between this thing and the frictional force in between this thing.

00:50 So that is the balance here.

00:53 It should be balancing.

00:54 So we can write down that m times a of the conveyor must be equal, must be balanced by, we can beforehand write force of the conveyor must be balanced by the tension in the string plus the frictional force in the conveyor plus frictional force of the box, between the box.

01:20 So we can write down tension, we can just write down t plus the frictional force will be static friction.

01:33 This is due to static friction between the conveyor, this is due to kinetic friction times the normal force which is the weight of the both the body.

01:43 So box two times the box, which is two times mass of the box times g, that is the kinetic friction plus static friction times mass of the box times g.

02:01 This is the balancing force.

02:02 So we can find out the acceleration.

02:05 So the mass times acceleration of this thing is actually it is holding the mass of two boxes.

02:14 Okay, that is what is happening.

02:15 So it is m1, two times mb times ac must be equal to t plus mu k times 2 mb g plus mu s times mb g.

02:32 So mb can be cancelled on both sides and you can say that ac must be equal to t by 2 plus mu k g plus mu s g by 2.

02:47 So we got the acceleration here, right? so when the thread is in place, when the tension is active, that this acceleration is going to be equal to zero, right? this will be balanced.

03:00 That means the tension will be equal to mu k g plus mu s g by 2 times 2.

03:11 That will be acceleration because this is zero with a minus sign.

03:16 That will be the tension.

03:19 So we got the tension.

03:20 So the tension is when the rope is cut, what happens is that the acceleration, this acceleration starts to take place, right? acceleration starts to be active and what happens is that this whole thing will start to move.

03:39 It will start to move.

03:41 So we can say that acceleration, that acceleration will be equal to tension, which is minus mu k g plus mu s g by 2, right? times 2 and 2 will cancel each other.

03:59 So plus mu k g plus mu s g by 2 will be active, right? and this minus sign goes to the inside as well.

04:09 This balances each other.

04:11 So the acceleration will be equal to zero when the tension is that, okay? that is, that is a balancing condition.

04:19 Now, now what we can do is that we can now define the acceleration.

04:23 So now the tension, when the, then the thread is cut in this system, when this is cut, the t disappears from this equation.

04:33 So acceleration will become mu k g plus mu s g by 2...