Block on a wedge connected to pulleys

[Hamiltonian]

Homework Statement

in the figure block moves downward with a velocity v1, the wedge moves rightwards with velocity v2. the correct relation between v1 and v2 is.

Relevant Equations

-

if the tiny block moves downward by an amount x, the wedge should also move forward by the same amount x as they are connected by the same string whose length has to remain constant, (by differentiating it wrt time we get speed) hence I concluded that v1 = v2, but my book says otherwise what is wrong with my intuition here.

 

[haruspex]

they are connected by the same string whose length has to remain constant,

Yes, but the distance between the two pulleys reduces, leaving more string elsewhere.

 

[phinds]

Looks to me like the pully from which the V1 block is hanging is tied to the wedge, which complicates things. I'm agreeing w/ haruspex.

 

[Hamiltonian]

Yes, but the distance between the two pulleys reduces, leaving more string elsewhere.

the distance between the pulleys reduce because the tiny block moves down

 

[haruspex]

the distance between the pulleys reduce because the tiny block moves down

No, it reduces because the wedge moves right.

 

[kuruman]

Homework Statement:: in the figure block moves downward with a velocity v1, the wedge moves rightwards with velocity v2. the correct relation between v1 and v2 is.

"The correct relation between v1 and v2 is $\dots$" is what? Are we looking at a multiple choice problem with missing choices?

In any case, I would write the total length of the string as the sum of three pieces, one horizontal, one vertical and one slanted. Note that angle $\theta$ depends on time which means that the length of the slanted piece also depends on time. That's what @phinds and @haruspex were alluding to.

 

[Hamiltonian]

"The correct relation between v1 and v2 is $\dots$" is what? Are we looking at a multiple choice problem with missing choices?

yes its a multiple choice problem

 

[kuruman]

yes its a multiple choice problem

And the choices are $\dots$?

 

[Hamiltonian]

a) v2 = v1
b) v2 = v1*sin(theta)
c) 2(v2)sin(theta) = v1
d) (v2)(1+sin(theta)) = v1

 

[haruspex]

a) v2 = v1
b) v2 = v1*sin(theta)
c) 2(v2)sin(theta) = v1
d) (v2)(1+sin(theta)) = v1

Consider two extremes, $\theta =0$ and $\theta \to\pi/2$.

 

[kuruman]

a) v2 = v1
b) v2 = v1*sin(theta)
c) 2(v2)sin(theta) = v1
d) (v2)(1+sin(theta)) = v1

Thank you for providing the choices. Strictly speaking they are all incorrect. The statement of the problem defines v1 and v2 as "velocities." The horizontal length of string is decreasing whilst the vertical length is increasing. Therefore there a negative sign is needed in front of v1 in all the expressions. Although the problem asks for a relation between velocities, the provided choices are relations between speeds.

One of the listed expressions provides the correct relation between speeds and you can identify it by following @haruspex's suggestion in #10. It would be instructive, however, to actually derive that expression. Then you will become more confident that you know how to "do" problems of this kind just in case you see one again on a test but without choices.

 

[Hamiltonian]

Consider two extremes, $\theta =0$ and $\theta \to\pi/2$.

using this method, I feel it should be option

d) (v2)(1+sin(theta)) = v1

as both the wedge and tiny block will move with the same speed when theta = 0, but if let's say there was another option (e) (v2)(1 + 2sin(theta)) = v1 there would be no way to discard this right?
also, you could also get confused with option b) v2 = v1*sin(theta) if you have the wrong intuition

 

[kuruman]

using this method, I feel it should be option

as both the wedge and tiny block will move with the same speed when theta = 0, but if let's say there was another option (e) (v2)(1 + 2sin(theta)) = v1 there would be no way to discard this right?
also, you could also get confused with option b) v2 = v1*sin(theta) if you have the wrong intuition

That is why it is safer to derive the correct expression. I have sometimes seen multiple choice questions with all answers incorrect because of some typo or error by the author. In other cases, more than one answer could be construed as correct because of ambiguities in the statement of the problem. Is your goal to learn how to do physics or how to eliminate the wrong answers assuming that only one of them is correct?

 

[Hamiltonian]

Is your goal to learn how to do physics or how to eliminate the wrong answers assuming that only one of them is correct?

my goal is to learn physics I was only responding to a method of solving a problem, as I felt that method isn't a very general method.

 

[kuruman]

my goal is to learn physics I was only responding to a method of solving a problem, as I felt that method isn't a very general method.

Great. If you decide to derive the correct answer, we are here to help.

 

[haruspex]

let's say there was another option (e) (v2)(1 + 2sin(theta)) = v1 there would be no way to discard this right?

No, that would not work in the $\pi/2$ extreme. The section of rope connecting the two pulleys approaches the horizontal, and it should be clear that v1 tends to 2v2, not 3v2.

The main reason I suggested to consider these cases was to make it clear that the horizontal distance between the pulleys (equivalently, the angle theta) affects the ratio of the two speeds. But it can also be handy in the time pressure of an exam.
As @kuruman says, the more instructive approach is to derive the relationship independently of the offered choices.

Create unknowns for the the lengths of the three sections of rope, for the total length of rope, and for the vertical height between the two pulleys. What two equations can you write relating those and the angle?