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Faris Shajahan
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Please help me with https://www.physicsforums.com/threads/i-need-an-answer-fast.799372/jedishrfu said:Welcome to PF!
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Please help me with https://www.physicsforums.com/threads/i-need-an-answer-fast.799372/jedishrfu said:Welcome to PF!
Well its not my homework...its a doubt...jedishrfu said:We don't work that way here. We can't do your homework for you. We can only provide hints once we see some work from you.
Please take some time to read the forum rules.
Faris Shajahan said:we take component of v1 along the string i.e. v1cosθ \theta and write v1cosθ \theta =v2.....but why don't we write v2cosθ \theta =v1?
Could you please explain a little more...because I did not get what you exactly meant!PeterDonis said:Because the angle ##\theta## is between the direction of v1 and the direction of the string at m1. It is not an angle associated with v2 or m2. The direction of v2 is the same as the direction of the string at m2, so there's no angle involved there.
You can.jeffery_winkle said:With enough pulleys, one person could lift ten tons.
Are you saying that the method that I outlined in post #11 will give the wrong answer? Because I don't think so.NascentOxygen said:In answer to the very first post: the component of v1 in the direction of the string is the method to use.
The right-angled triangle you see outlined by the string and the horizontal is not a velocity vector triangle, neither is it a distance vector triangle. While its base and hypotenuse can change, this triangle's height remains fixed by the location of the pulley, so it doesn't represent any vector triangle. You'll have to construct your own vector triangle for the analysis, and it won't look like the right-triangular shape formed by the string around the pulley. The latter is just a clever distraction to trip up the unwary.
Sorry. I started my response when there were only 2 other replies, and I was addressing post #1. Only after I'd posted did I become aware of still more replies, but I haven't studied them.Chestermiller said:Are you saying that the method that I outlined in post #11 will give the wrong answer?
I know what you mean and this method, I believe, uses constraints.Chestermiller said:Let h be the distance of the pulley above the table. In terms of h and θ, what is the horizontal component of distance of the ball from the pulley? What is the distance of the ball from the pulley along the hypotenuse? What is the time derivative of the distance along the hypotenuse, given that h is constant? What is the time derivative of the horizontal component of distance of the ball from the pulley, again given that h is constant? What is the relationship between these two time derivatives?
Chet
Sure. See post #16.Faris Shajahan said:I know what you mean and this method, I believe, uses constraints.
It goes like...
Let the length of the string between ##m_1## and the pulley at any time t be ##L##.
And let the distance between the pulley and the table be ##y##. Let the horizontal distance between ##m_1## and the pulley be ##x##.
Then as ##m_1## moves ##L## changes, ##x## changes, but ##y## remains constant.
Considering the right angled triangle,
##L^2=x^2+y^2##
Differentiating,
##2L\frac{dL}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}##
From the figure, ##\frac{dL}{dt}=v_2## (as with whatever velocity the string moves on the right side of the pulley, the same on the left) ##;\frac{dx}{dt}=v_1; \frac{dy}{dt}=0##
Also ##x=Lcos\theta##
Hence substituting,
##Lv_2=Lcos\theta v_1##
##=> v_2=v_1cos\theta##
Clear.!
But any other way than going for all this or in other words, how do we find this directly?
Clearly possible.jeffery_winkle said:This is a related question. I have heard it said that with a system of pulleys, it reduces the force you need to lift a given weight, so someone could lift something heavier than they could normally lift. Well, if that were true, by simply adding more pulleys, you could lift an unlimited amount of weight. With enough pulleys, one person could lift ten tons. Well, that can't possibly be true. What's the flaw in the reasoning?
Oh right, I hadn't seen that!Chestermiller said:Sure. See post #16.
Chet
The solution to this pulley problem depends on the specific details of the problem, but there are some general principles that can be applied. Here are five frequently asked questions about finding the solution to a pulley problem.
To find the tension in the rope, you will need to use the principle of conservation of energy. Start by identifying the pulleys and the masses involved in the problem. Then, use the equation T1 + T2 = 2m1g to solve for the tension in the rope.
In most cases, you can ignore the mass of the pulley when solving a pulley problem. The mass of the pulley will only have a significant effect on the solution if it is very large or if the rope is very thin and light.
If there are multiple pulleys involved in the problem, you will need to use the concept of mechanical advantage. The mechanical advantage of a pulley system is equal to the number of ropes supporting the load. Use this concept to determine the relationship between the tensions in the ropes and the masses involved.
If there is friction involved in the pulley problem, you will need to take it into account when calculating the tension in the rope. The presence of friction will cause the tension in the rope to be slightly less than what it would be in a frictionless scenario.
No, equations of motion (such as Newton's laws) are not typically used to solve pulley problems. Instead, the principles of static equilibrium and conservation of energy are more commonly used. However, if the problem involves a mass moving with acceleration, then you may need to use equations of motion to find the solution.