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Problems involving pulleys can seem difficult at first glance, but they don't have to be!
In this video we will learn how to take a complicated pulley problem like this one,
and simplify it into something much easier to work with.I
f this video helps you, please subscribe or share it with a classmate.
Here we have a large 10kg box on a frictionless surface. A 2kg mass is attached to the 10kg
mass by a pulley. This question requires us to find the acceleration of block A. Since
both blocks are attached together by a rope, they will both have the same acceleration.
If we draw the individual free body diagrams, we quickly see that both have a force of tension.
Since they are attached by the same rope, the tensions will be the same. I've drawn
this in purple do you can better visualize it.
Now we have some good news and some bad news. The bad news is that we don't know the force
of tension and without it we can't directly solve for acceleration. The good news is that
we can make this problem much simpler to solve.
All the rope in this problem was doing was changing the direction of the weight from
box B. In other words, the force of gravity from box B was being redirected
by the rope and pulley to move box A along the surface.
So let's forget about the tension for now, and instead create a different, but equivalent
problem that focuses on the forces that are actually doing the moving.
Just because the rope redirects the weight of B does NOT mean that the weight and tension
are equal. In fact, if the boxes are accelerating, they will not be equal.
OK. So now we see that the force of gravity from box B is what is providing the force
to move box A. Now that we see this, we can forget all about the entire pulley system
and just pretend the box is being pushed with that same 19.6 N.
Congratulations! We've converted a complicated problem to a very basic one! Reward yourself
by drawing a new free body diagram. You may find it beneficial to include the values for
your variables on the free body diagram. It's optional, but I'll do it here.
The next step is to apply newtons second law. The net force is equal to the total mass of
the system times its acceleration.
Since block A is not moving in the y direction, the net force there is zero, and the normal
force and the weight of A are balanced. We don't need this information in this problem
so we move on to finding the net force in the X direction.
Since both box A and B have mass, they have inertia that will affect the acceleration
of the system. Newton's second law requires us take the total mass of the system into
the equation.
Now to find the net force. Since we don't have friction or any other forces, the only
force we need to include is the force of gravity from mass B. If this problem had included
friction we would have to subtract that. Lucky for us we don't.
That's it for this problem! Just solve for acceleration and put the numbers in your calculator!
Give this problem a try on your own, except this time solve it when box A has a 5N force
of friction.