- #1
new90
- 90
- 9
- 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
You are wrong on both counts.new90 said: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
Perhaps you did not understand my question.new90 said:i think none but I think is wrong
It's easier than that, if you make a small approximation.new90 said:I think i can use the pythagorean theorem to find PQ' IM correcT?
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.new90 said:View attachment 261702
the letters are in correct place
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 clearlyharuspex said:Thanks @Adesh ... though it's really the OP's job to draw diagrams.
haruspex said:and P be the point where the string going up and left from there contacts the pulley.
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 said: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 :-)
I didn’t get that, do you mean the component of vertical velocity along PQ?haruspex said:what component of that is in the direction of the pulley?
Right. So can you see new90's error in post #1?Adesh said: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##
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 said:Right. So can you see new90's error in post #1?
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.Adesh said: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?
So, if A is descended instantaneously by ##y## the ##1~kg## mass will move up by ##y/2##.haruspex said: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.
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 said:THanks so much haruspex for your patience I know I am pretty dumb
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 .haruspex said: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?
Was there more to the question? If so what was the rest? Sorry for my ignorance!new90 said:i was wrong i don't know to solve it
I use the virtual work but the answer is not correct
Please post all the steps of your working, and not as an image.new90 said:sin of the angle times v thets what I used to the left side but i think is not the correct process
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.new90 said: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
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?new90 said: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