Part A In the block-and-tackle arrangement shown in (Figure...

Part A In the block-and-tackle arrangement shown in (Figure 1), three segments of the single rope pull on the block. What is the magnitude $F_{px}$ of the force exerted by a person on the rope to raise the block at constant speed? Express your answer in terms of the inertia of the block $m$ and the gravitational acceleration $g$.

Transcript

00:01 So this question has two parts.

00:05 So we'll take the first part on the right side.

00:08 Right side diagram shows that there are two blocks.

00:10 One small block is one kg in weight.

00:13 The bigger block is three kgs in weight.

00:16 And there is a five newton force acting towards the right on the smaller block.

00:23 And there's a three newton force acting towards the left on the bigger block.

00:27 So with this condition we can assume that there's no friction on the floor here.

00:31 And this this condition, we can assume that there's no friction on the floor here.

00:31 And this, this condition, we are asked to find what is the other few things we have to find the first part is what is the acceleration of this system both the blocks together so since we know that this is a bigger force on the system than this force if we assume this to be as one mass so we can assume that there is going to be an acceleration in this direction it's called that acceleration a so with this we can draw the free body diagrams for both these cases so let's say that this is small box a block sorry and this is the bigger block so there is a 5 newton force here and there is a 3 newton force here sorry let's also assume that there is a force between these blocks which will be equal and opposite let's call that f and that's because if one body applies f to the second body the second body in reaction applies f back that's new is third law equal and opposite reaction.

01:40 So given that this is moving at acceleration a, there'll be a pseudo force which will be 1 kg times a and since this is also moving at acceleration a, there'll be a pseudo force three times a.

01:55 So now we can write the equation for both these cases.

01:58 So for the smaller block let's say for the 1kg block it says that 5 newton in the right direction is equal to f plus a or one times a in the left direction and for the three kg case we can say that the three towards the left 3 newton force was the left is equal to f newton's towards right and minus 3 a towards the right as well since 3a is in the left i can say minus 3 a towards right or i could have also written this as f is equal to 3 plus 3a.

02:44 So i have basically just rearranged this equation, this equation.

02:50 And why i have done this is because once i get in this form, i can subtract these equations first.

02:56 And i subtract these equations, i get 2 is equal to 0 because f gets cancelled out.

03:05 So 0 and plus 4a.

03:08 So i am basically subtracting 3 from 1.

03:11 So it will be 5 minus 3 and a minus 3a which is because becomes 4a.

03:16 So from here i can say that a is equal to 2 by 4 or 0 .5 meters per second square.

03:25 And this is the answer for your first part.

03:27 This is the acceleration of the system under the influence of these two forces.

03:32 Okay...

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