Home/Archive/Archive - 2015-16/Physics 03, Pulley Block Problems

## Physics 03, Pulley Block Problems

Okay, we have the questions, I will tell you the answers. Have you reviewed the lectures?
Are you telling the truth? Kids, make a permanent habit of this, such small things are very
useful.
First part was, when the man is at rest means T must be equal to Mg. In fact, we will make
free body diagram. This year man and string diagram will be used frequently. Many
questions are solved by this free body diagrams. T and Mg. See, stand in the place of this
man. Suppose you have to hang to a string, what will you do? You will pull string
downwards, right. You will pull the string downwards, so the string will pull you upwards.
Suppose we see tension force on the string, then it will be downwards and if we see tension
force on man, it will be upwards. So, keep this in mind it is always true. And if the man is at
rest than must be T equal to Mg.
Second case, if man is moving upwards with constant speed, constant velocity means
acceleration is zero. So, what will be the answer of part b? T is equal to Mg. Has everybody
done this question?
Third, moving down with constant speed means, down with constant speed T is equal to Mg.
Moving upwards with constant acceleration means T minus Mg is equal to Ma, yes or no. So,
effectively T is M into g plus a. Man is moving down with constant acceleration, so, it will
become Mg minus T equal Ma. So T will be M into g minus a, okay. How many of you have
done it correctly, everyone, okay?
Last example, Pulley block. This is a popular variety, it will come of use many times, see it
carefully. Find the acceleration of the blocks. Let’s see the first part of the question. Find the
acceleration of the block. Okay, calculate. Okay, then what should our first step be, kids?
Free body diagram. Okay, let’s make our first step clear. 100 Newton, Tension, Tension 50
Newton. Is it clear, kids? Next, step will be to assign acceleration, so, 10kg will come down
and 5kg will go up. Done, any doubts, okay. Let’s write the resultant force on 10kg and that
should be mass into acceleration, any doubt. For 5kg this equation will be. Now we have two
equations in two variables. Add them, how much, 50 is equal to 15a. a will be 50 by 15, 10 by
3, done. Everything is clear, each and everyone. Now, we will proceed further, each and
every students except new students, other students answer should be correct, done. Now,
we will take part b of this question. Find the force applied by string, force applied by string
on the pulley, string on the pulley. Students, see here, I will do it. We will make free body
diagram of this pulley. We will make free body diagram of everything. Now, the pulley is
massless. Stop writing, listen carefully, other examples you have to do. I will just do this one.
This is a pulley, FBD will be made of this. Now, pulley is mass less, so there is no role of Mg.
what is our clarity WTN. There is no W, if pulley had a weight, we would have represented
that also. Now, tension, I will take these strings. Now, if we see carefully, that this string is
pulling the pulley from both the sides, here also T and there also T. It is a rule that, if a string
passes over the pulley, that string will always pull the pulley with the help of two tension
force. It is not possible that there is only one tension available and the other tension is not
available. So, we have to make this TNT, the string which is moving over the pulley, that
string will always apply two tension forces and both of them will be of any angle, it can be
parallel or 90 degree, it can be anything. It depends on the situation, got my point. The string
which passes over the pulley will apply tension force in two directions, got my point. Two
directions means there will be two different tension forces on the pulley due to strings.
Now, as it is on same direction then both the force vector will be added and if it was in 90
degree, then we would have added it in vector way, root of a square plus b. So, in that
scenario, it is two T down. Now, listen carefully, the examiner will ask it two three questions,
listen carefully. Now, this string is there, if we see ideally, this string is holding the pulley.
This can be a string, rod or anything else. If I make a complete free body diagram of the
pulley, so will I get one more force here, yes or no, T dash. So this is the complete free body
diagram. Who applied this T dash, this one string and who applied this T and T, these both
strings. And see, we can see that the pulley is at rest, if anything is at rest, what will be its
total force, zero. So, I can say that T dash is equal to 2T and from where I can get the value
of T, from here. But what was important for me, to know that the pulley which passes over
the pulley, that pulley always has tension force in two directions or tension force is applied
in two ways. Done, everyone, any doubts, anyone. Next question. Copy it down. Pulley is at
rest, so the total force is zero then the upper force is equal to the force below. Which one,
these two, they are only one string. Mass less in extensible string, the tension in full string
will be the same. See here, this is 10 and 50, because of this both, effectively, this T will be.
Wait, I will show a magic, listen carefully. We will calculate here. 50 plus 5 into a, 50 plus 5
into 10 by 3. So, effectively, what will happen 200. Tension is 200 by 3. Now, understand this
take the 10kg block, 100 down and 200 by 3 is up, so, what will be the resultant force of
this? 100 minus 200 by 3, how much it is, 100 by 3, net force divided by its mass, what will
you get acceleration, okay. Now, take 5kg block, 200 by 3, 50, resultant force will come up,
how much, 200 by 3 minus 50. How much? 50 by 3 divided by its mass, what is the
acceleration 10 by 3. So, what is the condition, the condition is that, both the acceleration
should be same. Now see this, in this 100 is applied and on this 50 is applied. By seeing 100
and 500, the tension of string selects one unique value, that unique value on both of these
blocks will generates that much resultant force, that makes the acceleration of both the
blocks same. So, tension will be same because there is only one string there will be no
individual impact of 10kg and 50kg, it makes a net effect, what happens with the net effect,
we will select a middle tension, that is 200 by 3, which is 66 point 7. We selected a unique
value of the tension which was greater than 100 and smaller than 50, by which the
acceleration of both will effectively be same. So, there is no role of tension if mass is
separate. Tension selects its unique value. And secondly we had made a rule, that the
tension of one in- extensible mass less string will always be the same. Even if you connect
any block or you don’t connect, it will not make any difference. Okay, next question,
understood kids, I didn’t go fast, sure, everyone.
Okay, find the acceleration of the block and tension force in A, B string. Find the acceleration
of the block and tension of force in A, B string. You have not placed blocks, but still the
tension will be the same because that is the property of the massless inextensible string. 6
kg block and 60 Newton force, they are two different things. Okay, then listen very carefully.
We will solve two problems then we will compare both of them. Then we will come to know,
what is the difference between them? Listen very carefully, I had told you this thing in the
class yesterday. In any string, information means you are pulling this end of string
downwards with 60 Newton. That means tension of this string is, 60 Newton. So, in this full
string how much tension force exists, 60 Newton. This is what we had discussed yesterday.
Now, we will make FBD of 4kg, 40 and tension, how much, 60 Newton. How many of you
knew that tension will be 60 Newton? Has the force not increased? Kids, grow up, keep
some basics things with you. See, in the last lecture also, I had made like this circle and star
and told you this is a special thing, right. Okay, listen, if you see resultant force on 4kg, 60
minus 40 is equal to 4a, so, 20 is equal to 4a. So, a is equal to 5 meter per second square. a is
equal to 5 meter per second square and tension is 60Newton. But, I asked you one more
question, tension of A, B strings. If I want know the tension of A, B, then I will have to make
free body diagram of pulley. So, from here also 60 Newton and there also 60 Newton.

2016-06-13T14:15:42+00:00 Categories: Archive - 2015-16||0 Comments
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