﻿ BCLN - Physics - Dynamics - Pulley Problem
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# BCLN - Physics - Dynamics - Pulley Problem

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After learning about FBD's and Fnet = ma, you can learn about combined-mass problems. This is the second combined-mass example. It involves two masses and a pulley (also called an Atwood machine).

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0:00now that we know how to deal with connected masses or systems of masses
0:05we can also solve pulley problems
0:09the same general ideas can be used as well see for example two masses 100
0:17kilograms and fifty kilograms are at each end of a rope that is hung over a
0:23frictionless pulley determine the tension in the Rope again we recognize
0:29that there are three possible free body diagrams here
0:33the first one the 100 kilogram mass we have FG pointing down the force of
0:40gravity and T the force of tension from the rope pulling up again we know that
0:46the Rope is pulling up in that
0:48well if we cut the rope that mass would definitely fall more quickly now before
0:53we get to our equation
0:56let's determine our positive direction again it's up to you to choose the
1:01positive direction but to keep things simple it's always nice to make the
1:05positive direction in the direction of motion since it'll move in the direction
1:10of the heavier mass to the left
1:13we will consider down as the positive for this equation therefore we'll have f
1:18net equals m1 a and F net is fg1 the force of gravity pulling it down and
1:26minus T the tension and the rope pulling it back up and that equals m1 a and we
1:34know the mass m and can determine the force of gravity that is mg but we still
1:41have two unknowns tea and a
1:46next we can consider the 50 kilogram mass so in this one we have again
1:53FG pulling down the force of gravity and teeth they tension force from the rope
1:59pulling up the force from the Rope is equal and opposite to the force acting
2:04on mass one so calling them both T is going to be helpful
2:09now before we get to our equation let's again determine the positive direction
2:14for this one
2:16now in this case positive would be up again we're looking for the direction of
2:22the expected motion
2:23it may seem weird but it's going to be the same direction in fact as we
2:29consider how the positive in the other free body diagram is down and the Rope
2:34is being pulled over pulley
2:36even though one is up and the other is down there both consistent
2:41the system will drop on the left side and raised on the right side and our
2:47equation would be F net equals m2 a and in this case from the positive direction
2:53we have tea
2:54pulling it up and the FG to pulling it back down and that's equal to her
3:01m2a again we know the mass m and can determine the FG - that is mg but we
3:12still have two unknowns the tension and the acceleration
3:18our third free body diagram will be the entire system but before we start
3:25here's a little trick that'll make things easier for you for these pulley
3:29problems you know how down was the positive direction on the Left mass and
3:35then we have to consider up as the positive direction on the right-hand
3:39mass as this is the direction
3:42each will move now this is due to considering a pulley
3:48we're going up on one side and down on the other well is just how it fully
3:53works well the difficulty is that considering positive to be down on the
3:59left and up on the right
4:01gets a bit confusing when you try and combine them as an entire system which
4:07should be positive and which should be negative
4:10well we can simplify this by drying or free body diagram as if we kind of
4:15flattened out the rope so we consider left to be positive in this case and
4:20right to be negative and it's all consistent same forces involved fg1 and
4:26FG to pulling in opposite directions
4:29just like on the pulley therefore we simply have an easier way to consider
4:35this same situation
4:37it's totally consistent with the real situation our equations would be F net
4:43equals MA
4:44again will show the M as m1 plus m2 to remind us that we are considering the
4:52entire system for this equation f g one positive because it's going to the left
4:58and FG to would be negative because it's pulling to the right and that's equal to
5:06m1 plus m2 the total mass times the acceleration but now we know enough that
5:12we can rearrange and solve for a
5:18and now that we have
5:19the acceleration of the system we can solve for the row tension T using either
5:25one of the other free body diagrams
5:27so let's try using the hundred kilogram free body diagram we already know that
5:33ft 1 minus T equals m1 a so let's just rearranged 40 and plug in our
5:40acceleration again all the items in the system moved together and accelerate at
5:45the same rate and we can solve for T
5:51and we could stop there but let's verify our answer by considering the mass to
5:57free body diagram again by Newton's third law the tension T on the 50
6:03kilogram mass had better be equal and opposite to the tension on the 100
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Pulleys with scale, pulleys, how pulleys work, physics