Three masses connected by pulleys

[new90]

Homework Statement

i Dont know why he uses sin-1c i need to find displacement

Relevant Equations

no equation

I know that if A moves down 1 unit b will move up square root of 2 and 1kg will moves up 2 unit

 

[haruspex]

Homework Statement:: i Dont know why he uses sin-1c i need to find displacement
Relevant Equations:: no equation

View attachment 261508

I know that if A moves down 1 unit b will move up square root of 2 and 1kg will moves up 2 unit

You are wrong on both counts.
Draw a diagram showing the string from left hand pulley down to the junction above A. Draw it again with A a little bit lower. What is the change in length of that section of string?

 

[new90]

i think none but I think is wrong

 

[haruspex]

i think none but I think is wrong

Perhaps you did not understand my question.

Let Q be the point where the strings join above A, and P be the point where the string going up and left from there contacts the pulley.
If A is pulled down a distance y, the junction of the strings moves down y to a new point, Q'.
How much longer is PQ' than PQ?

 

[new90]

I think i can use the pythagorean theorem to find PQ' IM correcT?

 

[haruspex]

I think i can use the pythagorean theorem to find PQ' IM correcT?

It's easier than that, if you make a small approximation.
Drop a perpendicular from Q to the line PQ', meeting it at R. If y (the distance QQ') is small then the distances PQ and PR are near enough equal. So the increase in distance from P is RQ'.
What is the angle QQ'R? So in terms of that angle and y, what is the distance RQ'?

 

[new90]

the letters are in correct place

 

[haruspex]

View attachment 261702
the letters are in correct place

Yes, except that I thought you phrased your question in terms of what happens to the 1kg mass, so P should be at the left hand pulley.

Anyway, you don't need to draw the whole diagram. Just draw the triangle PQQ', quite large, then drop a perpendicular from Q to PQ' to meet it at R.
The length of string from the pulley to the junction is PQ to begin with, PQ' after A has moved down. PQ is very nearly the same length as PR, so the increase in length of that section of string is RQ'.
What is the equation relating that increase to y (=QQ') and the angle PQ makes to the horizontal?

 

[Adesh]

@new90 haruspex wants you to make and refer to these images:

Image 1
Image 2
Image 3

Refer to these images and read post #8 carefully.

 

[haruspex]

@new90 haruspex wants you to make and refer to these images:

Image 1
Image 2
Image 3

Refer to these images and read post #8 carefully.

Thanks @Adesh ... though it's really the OP's job to draw diagrams.

 

[Adesh]

Thanks @Adesh ... though it's really the OP's job to draw diagrams.

Sorry sir, I thought OP was feeling the same thing which I feel, whenever I get a doubt I think too much that my thinking ability gets impaired and even the simplest of things can’t be understood by me in that state of mind. I could see OP made an unusual error in denoting when you said it very clearly

and P be the point where the string going up and left from there contacts the pulley.

By the way I too need the solution of this problem :-)

 

[haruspex]

Sorry sir, I thought OP was feeling the same thing which I feel, whenever I get a doubt I think too much that my thinking ability gets impaired and even the simplest of things can’t be understood by me in that state of mind. I could see OP made an unusual error in denoting when you said it very clearlyBy the way I too need the solution of this problem :-)

You might find it easier to think in terms of velocity components. If the joint above A moves vertically with velocity v, what component of that is in the direction of the pulley?

 

[Adesh]

what component of that is in the direction of the pulley?

I didn’t get that, do you mean the component of vertical velocity along PQ?

The component of vertical velocity $v$ along PQ is $v/2$

 

[haruspex]

I didn’t get that, do you mean the component of vertical velocity along PQ?

The component of vertical velocity $v$ along PQ is $v/2$

Right. So can you see new90's error in post #1?

 

[Adesh]

Right. So can you see new90's error in post #1?

If in one second A moves by $v$ units downwards, the 1 kg block will up by $v/2$ units in that one second. Am I right?

 

[haruspex]

If in one second A moves by $v$ units downwards, the 1 kg block will up by $v/2$ units in that one second. Am I right?

Yes, except that one second may be rather long. As soon as the angle changes the ratio will change, so it's more a statement about the instantaneous velocity ratio.

Note the guess in post #1 had it backwards, with the 1kg mass rising at twice the rate A descends. This is a common mistake.

 

[Adesh]

Yes, except that one second may be rather long. As soon as the angle changes the ratio will change, so it's more a statement about the instantaneous velocity ratio.

So, if A is descended instantaneously by $y$ the $1~kg$ mass will move up by $y/2$.

And similarly, any instantaneous descend of A by an amount of $y$ would result in $\frac{y}{\sqrt 2}$ ascend of block B.

Thank you so much sir.

 

[new90]

I dot understand why y/2 and y/square root of 2

 

[new90]

sorry I understand now

 

[new90]

THanks so much haruspex for your patience I know I am pretty dumb

 

[Adesh]

THanks so much haruspex for your patience I know I am pretty dumb

Never tell yourself you’re dumb, it was really a hard problem. There are many people out there to discourage you so you don’t need to discourage yourself, leave it to them.

 

[new90]

thanks

 

[new90]

i was wrong i don't know to solve it
I use the virtual work but the answer is not correct

 

[etotheipi]

You might find it easier to think in terms of velocity components. If the joint above A moves vertically with velocity v, what component of that is in the direction of the pulley?

I just did it by saying that if $\delta t$ is sufficiently small, PQ and PQ' approach the same length and the triangle PQQ' becomes isosceles. So you know the other two have to take up the slack of $\frac{v\delta{t}}{2}$ and $\frac{v\delta{t}}{\sqrt{2}}$ respectively. I realized it's essentially the same approach you took, except yours in terms of components of velocity was much nicer .

i was wrong i don't know to solve it
I use the virtual work but the answer is not correct

Was there more to the question? If so what was the rest? Sorry for my ignorance!

 

[new90]

i don't understand how you get the square root of 2

 

[etotheipi]

OK; with @haruspex's method you need to find the component of the velocity in the direction of the string on right. If the velocity is $v$ and points vertically upward, and the right string is at $45^o$ to the velocity, then what is this component?

 

[new90]

sin of the angle times v thets what I used to the left side but i think is not the correct process

 

[haruspex]

sin of the angle times v thets what I used to the left side but i think is not the correct process

Please post all the steps of your working, and not as an image.
.

 

[new90]

so what i do was that if A drops 1 units |kg drops 1/2U then i used the the virtual work and i canceled y in both sides this is equal to 10N times 1/2 =10A and that equal to .5 kg wt

 

[jbriggs444]

so what i do was that if A drops 1 units |kg drops 1/2U then i used the the virtual work and i canceled y in both sides this is equal to 10N times 1/2 =10A and that equal to .5 kg wt

When struggling to understand something it is worthwhile to talk slowly. Show more detail in your work. Fill in all of the missing pieces. Dot all the i's and cross all the t's.

Don't say "i used the virtual work". Show us how you used virtual work.
Don't say that you canceled the y on both sides. Show us the equation where you canceled the y.

And try to avoid inconsistent abbreviations. Don't say "units" in one place and "U" in another.

 

[new90]

ok so what I did was follw what harupex said about the distance = if A moves down 1 units 1kg box will move up 0.5 unit then i used the vitual work method =Σfi times virtual distance = 0 this is =10N(because the force of gravity)times .5unit - 10aN(a = the box of the center)times 1 unit = 0(is minus because the forces are in different directions)this equal to = 5N =10aNewtons and the answer is 5N/10N = a = .5N and the final answer is 5 kg-wt but the answer in the book is 1.4kg wt

 

[haruspex]

ok so what I did was follw what harupex said about the distance = if A moves down 1 units 1kg box will move up 0.5 unit then i used the vitual work method =Σfi times virtual distance = 0 this is =10N(because the force of gravity)times .5unit - 10aN(a = the box of the center)times 1 unit = 0(is minus because the forces are in different directions)this equal to = 5N =10aNewtons and the answer is 5N/10N = a = .5N and the final answer is 5 kg-wt but the answer in the book is 1.4kg wt

I cannot find in this thread a complete statement of the problem, so I don't know what you are trying to calculate. Are you trying to find the weights A and B for equilibrium?

I note you make no reference there to box B. That should feature in your virtual work calculation.
In order to solve the problem you will also need to consider horizontal equilibrium.

 

[new90]

ok i will make another tread

 

[jbriggs444]

If you are making another thread to start over on this problem, that is the wrong way to go about it. Deal with the shortcomings in this thread by adding content here.

If you are making another thread to start a new problem and are giving up on this problem, don't advertise your new thread here. That's tacky. Just tell us you are giving up on the problem.

 

[new90]

the system is in static equilibrium .Use he principle of virtual work to find the weight A and B.Neglect the weight of the strings and the friction in the pulley