What is the solution to this pulley problem?

[Faris Shajahan]

Welcome to PF!

Please help me with https://www.physicsforums.com/threads/i-need-an-answer-fast.799372/

 

[jedishrfu]

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]

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.

Well its not my homework...its a doubt...
I know how to solve the problem completely but there is a doubt I have which is killing me!

 

[jedishrfu]

Then why did you post it needing an immediate answer?

 

[jedishrfu]

What is the doubt? Perhaps that can lead to a better understanding.

 

[Faris Shajahan]

The doubt goes like while solving problems like the one shown in the figure, we take component of v1 along the string i.e. v1cos$\theta$ and write v1cos$\theta$=v2.....but why don't we write v2cos$\theta$=v1?

 

[jedishrfu]

It looks like in this example the string can't stretch and the pulley has no mass (hence no torque).

The v1cos(theta) is the projection of the speed along the string caused by the ball moving in the v1 direction. Since the string is inextensible (i.e. no stretching) then the v1cos(theta) speed is directly related to the speed of the block moving upward.

I can see the confusion you're having here If you were to do it starting with v2 then v2 would be along the string and hence the projection of v2 on the plane where the ball is moving is v2cos(theta).

I just can't see the reason of why one is chosen over the other. I still need to ponder this.

Someone else here at PF may have a better explanation.

 

[PeterDonis]

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?

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.

 

[Faris Shajahan]

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.

Could you please explain a little more...because I did not get what you exactly meant!
Thanks though, it cleared "some" parts of my doubt...

 

[jbriggs444]

Saying it in slightly different words...

The motion of m1 can be seen has having two components. One in the direction of the string (i.e. at angle $\theta$ below the horizontal) and one at right angles to that (i.e at angle $\theta$ left of vertical).

Since component in the left-of-vertical direction is at right angles to the string, it has no effect on the motion of m2. It acts to "unwind" some string from the pulley without actually causing the pulley to rotate. The component in the below-horizontal direction does have a direct effect on the motion of m2.

So we separate the leftward motion of m2 into those two components. The left-of-vertical component is equal to v1 sin $\theta$ and is discarded. The below-horizontal component is equal to v1 cos $\theta$ and is the important piece.

 

[Chestermiller]

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

 

[mfb]

Another approach:
$v_1 \cos \theta = v_2 \cos \phi$ where $\phi$ is the angle between the direction of motion of object 2 and the string. This angle happens to be zero here, so the second cosine term is 1 and can be ignored. $v_2 \cos 0 = v_2 \cdot 1 = v_2$.

 

[NascentOxygen]

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.

 

[jeffery_winkle]

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?

 

[mfb]

With enough pulleys, one person could lift ten tons.

You can.
But the distance you have to pull the rope to lift the weight increases with the same factor as the weight you lift, so you need the same energy.
Please start a new topic if you want to discuss that in more detail.

 

[Chestermiller]

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.

Are you saying that the method that I outlined in post #11 will give the wrong answer? Because I don't think so.

H = hypotenuse
L = horizontal length

$$H = \frac{h}{sinθ}$$
$$\frac{dH}{dt}=v_2=-\frac{h}{sin^2θ}cosθ\frac{dθ}{dt}$$

$$L=\frac{h}{tanθ}$$
$$\frac{dL}{dt}=v_1=-\frac{h}{tan^2θ}sec^2θ\frac{dθ}{dt}=-\frac{h}{sin^2θ}\frac{dθ}{dt}$$

So, $v_2 = v_1cosθ$

Regarding the loss of the triangular shape formed by the string around the pulley, I can make the pulley as small as I wish relative the H and L.

Chet

 

[NascentOxygen]

Are you saying that the method that I outlined in post #11 will give the wrong answer?

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.

Keep up your good work, Chet!

 

[Faris Shajahan]

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

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?

 

[Chestermiller]

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?

Sure. See post #16.

Chet

 

[Faris Shajahan]

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?

Clearly possible.
An example may be as shown,
Let T be the tension in the string joining m and the closest pulley.

Then in this figure we require a force>=T...where T=mg to lift the mass.
But here,

though T=mg, we only require force>=T/2 to lift the mass m!

 

[Faris Shajahan]

Sure. See post #16.

Chet

Oh right, I hadn't seen that!