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Velocity of sliding ladder

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A ladder of length L = 3 m is leaning against the wall and the angle between ground and and ladder is α=60°. The bottom of the ladder slides at speed Vb = 0.2 m/s. Find the velocity of top of the ladder in respect to
    1) wall;
    2) bottom of the ladder.


    2. Relevant equations

    vx = [itex]\frac{dx}{dt}[/itex]
    vy = [itex]\frac{dy}{dt}[/itex]

    3. The attempt at a solution

    tan(α) = [itex]\frac{y}{x}[/itex], so y(x) = xtan(α). Then we find the derivatives of both sides of equation:
    [itex]\frac{dy}{dt}[/itex] =[itex]\frac{d}{dt}xtan(α)[/itex]=>vy1=vx*tan(α)
    And so vy1 = 0.2*tan(60°)=0.34 m/s.
    So this is the velocity of top part in respect to wall, right? I'm not sure what to do next - to find the velocity in respect to the bottom. Are my calculations right?
     
  2. jcsd
  3. Dec 7, 2013 #2
    In the first part, you got the correct magnitude for the velocity, but not the correct direction. Don't forget that velocity is a vector. For the second part, express the velocities of the ends of the ladder in vector form relative to a fixed observer. Then to get the relative velocity of one end with respect to (i.e., relative to) the other end, subtract the vector velocity of one end from the vector velocity of the other end.
     
  4. Dec 7, 2013 #3
    So you mean that vy1 < 0? But what if I choose my Y-axis to be downwards?
    So let's say for the second part we choose the fixed observer to be the point of intersection of wall and ground. Vx then would be [itex]\vec{v}[/itex]x = x*t and [itex]\vec{v}[/itex]y=y*t ?
     
  5. Dec 7, 2013 #4
    Suppose you call Vb the horizontal speed of the bottom of the ladder (0.2m/s), and you call Vt the vertical speed of the top of the ladder (0.34 m/s), and, if you choose the y-axis as downward, then, relative to a fixed observer, the velocity of the bottom of the ladder is [itex]\vec{v_b}=V_b\vec{i_x}[/itex], and the velocity of the top of the ladder relative to a fixed observer is [itex]\vec{v_t}=V_t\vec{i_y}[/itex], where [itex]\vec{i_x}[/itex] and [itex]\vec{i_y}[/itex] are unit vectors in the coordinate directions. Since the wall is fixed, the velocity of the top of the ladder with respect to the wall is [itex]\vec{v_t}=V_t\vec{i_y}[/itex]. The velocity of the top of the ladder with respect to an observer traveling along with the bottom of the ladder is [itex]\vec{v_t}-\vec{v_b}=V_t\vec{i_y}-V_b\vec{i_x}[/itex].
     
  6. Dec 8, 2013 #5
    Got it, thanks!
     
  7. Dec 8, 2013 #6

    haruspex

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    The first part can be done without calculus. Just consider the component of each end's velocity parallel to the ladder. Since the ladder does not change length, what is the relative velocity in that direction?
     
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