1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Doubt in Newton's laws

  1. Feb 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Capture.JPG

    2. Relevant equations
    None....
    Newton's laws

    3. 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## ???
     
  2. jcsd
  3. Feb 23, 2015 #2

    Quantum Defect

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. Feb 23, 2015 #3
    Thanks! But what is wrong when we write ## v_2cos\theta=v_1 ##??
     
  5. Feb 23, 2015 #4

    Quantum Defect

    User Avatar
    Homework Helper
    Gold Member

    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
     
  6. Feb 23, 2015 #5

    Suraj M

    User Avatar
    Gold Member

    ##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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Doubt in Newton's laws
  1. Newton's law (Replies: 8)

  2. Newtons laws (Replies: 3)

  3. Newton's laws (Replies: 5)

  4. Newton's Law (Replies: 2)

  5. Newtons law (Replies: 5)

Loading...