32 in B 24 in T? D 32 in 28 in 10 in A W ?W=5.5 lb Two cords...

32 in B 24 in T? D 32 in 28 in 10 in A W ?W=5.5 lb Two cords are used to suspend a block in equilibrium in the manner shown. (ADB is a single continuous cord that passes around the small frictionless pulley shown at O. CD is a separate cord that attaches to C, and to the center of the pulley.) Because the pulley is frictionless, the same tension (magnitude) $T_1$ exists in both sections AD and BD. The tension in cord CD, $T_2$, is different. Find the two tension magnitudes $T_1$ and $T_2$.

Transcript

00:01 So we have three blocks on this table pulley system where the mass a block is 6 kilograms.

00:14 The mass b block is 8 kilograms.

00:21 And the mass c block is 10 kilograms.

00:26 And so let me draw what this pulley system looks like.

00:31 This is mass b, and it is attached by the same string to mass a and mass b.

00:55 And it's sitting, mass b is sitting on a table.

01:05 These pulleys are at the corners of the table here.

01:10 And so our question is asking, when the blocks are released, what's attention in this cord over here, right? so if we were to draw our forces on this, we have the force of gravity pulling, this should be mass c.

01:32 We have the force of gravity acting on mass c.

01:38 So i put f sub g sub c, g for gravity, and then we have the force of gravity, f sub a, acting on mass a and then we have a tension force here, tension sub a, and another tension force here, tension subc, we'll call it.

02:01 And then we also have the force of gravity on the middle block, but it's not going to matter, and you'll see why in just a second.

02:10 So what we want to do is create a system of equations of these three blocks to try and solve for the acceleration of the total system.

02:23 Because once we get the acceleration, we can solve for what the tension is on the right part of this system.

02:34 So let's go ahead and write out our equations for this.

02:38 I will call this equation one, and it's going to be for block a.

02:44 Where the tension a minus the force of gravity on a is going to equal mass times acceleration for block a.

03:03 And we can rewrite this for the tension, and that'll just be the tension on a is equal to the mass of a times the acceleration, plus the force of gravity on a.

03:23 So there we have equation one.

03:25 Equation two is going to be the force of gravity.

03:33 So we're assuming that once the system starts accelerating, it's going to go in this direction, right? because this is the heaviest block.

03:44 That's an assumption.

03:46 Therefore, we're taking this way to be, positive.

03:50 So in this case, it's going to be the force of gravity on c minus the tension on the strain attached to mass c is equal to mass c times acceleration.

04:10 Again, we can solve for the tension on c just to make everything a little neater.

04:15 And that's going to be the force of gravity on c minus the mass of c times the acceleration.

04:26 All right.

04:27 And now for the middle block, our equation number three is going to be, again, since everything is traveling this way, we have the tension on string c minus the tension on string c minus the tension of on string a is going to be equal to the mass of block b times the acceleration.

04:59 All right, so now that we've got a decent system of equations here to work with, we can just start plugging one thing into the other to get a nice and clean equation we can work with to solve for the acceleration.

05:17 So we know the force of gravity on all three blocks, and we know the masses of all three blocks.

05:31 So in this case, we want an equation where we can isolate the acceleration, and then we can isolate the masses and the forces of gravity all onto the other side of the equation.

05:43 So let's rearrange these equations and see what we can get.

05:49 Let me finish rearranging this here.

05:53 We have the mass of block b times the acceleration plus the tension on me...

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