Why do we take component of v1 along the string in Newton's laws homework?

[Faris Shajahan]

Homework Statement

Homework Equations

None...
Newton's laws

The Attempt at a Solution

Not the attempt, the entire solution is:
String length is constant...hence if on the right side of the pulley, the string goes up by x distance then on the right side also the string goes down by x distance. Differentiating, we get velocity of string is equal on both sides. Now taking component of $v_1$ along the string we get $v_1cos\theta=v_2$!

But my doubt is instead, why don't we take component of $v_2$ along the plank and write $v_2cos\theta=v_1$ ?

 

[Quantum Defect]

Homework Statement

View attachment 79511

Homework Equations

None...
Newton's laws

The Attempt at a Solution

Not the attempt, the entire solution is:
String length is constant...hence if on the right side of the pulley, the string goes up by x distance then on the right side also the string goes down by x distance. Differentiating, we get velocity of string is equal on both sides. Now taking component of $v_1$ along the string we get $v_1cos\theta=v_2$!

But my doubt is instead, why don't we take component of $v_2$ along the plank and write $v_2cos\theta=v_1$ ?

Since the length of string is a constant. The decrease in length on the right side, must be equal to the increase in length on the left side.

You want to know the rate at which the hypotenuse of the right triangle on the left side is increasing, for a constant velocity in the x direction.

I.e. L = SQRT (x^2 + y^2). If v1 = dx/dt, dy/dt = 0, what is dL/dt? This (dL/dt) must be equal to v2.

 

[Faris Shajahan]

Since the length of string is a constant. The decrease in length on the right side, must be equal to the increase in length on the left side.

You want to know the rate at which the hypotenuse of the right triangle on the left side is increasing, for a constant velocity in the x direction.

I.e. L = SQRT (x^2 + y^2). If v1 = dx/dt, dy/dt = 0, what is dL/dt? This (dL/dt) must be equal to v2.

Thanks! But what is wrong when we write $v_2cos\theta=v_1$??

 

[Quantum Defect]

Thanks! But what is wrong when we write $v_2cos\theta=v_1$??

L = sqrt(x^2 + y^2)

dL/dt = 1/2* 1/sqrt(x^2 + y^2) * (2x*dx/dt + 2y*dy/dt) [dx/dt = v1, dy/dt = 0] ==> dL/dt = [x/sqrt(x^2 _ y^2)]*v1 = cos(theta) * v1

When you differentiate L (above) w.r.t. time you get dL/dt = cos(theta)*v1. Setting this equal to v2 gives you: v2 = cos(theta) * v1

Another way to think about this is that all of v2 goes into shortening the length of the string on the right side, but you need to have the ball move faster than v2 to take up the slack that is produced by the shortening of the right side. I.e. v1 = v2/cos(theta) ==> v1> v2

 

[Suraj M]

$v_2\cos(\theta)$ would give you the horizontal component of v₂ but you need the component of the velocity of the ball along the rope, not the same thing!
as Quantum defect pointed out if you say $v_2\cos\theta = v_1$ $\cos\theta =[-1,+1]$ so v₂will always be more than v₁by this equation which is wrong.