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Hypothetical kinetics questions, just for fun

  1. Mar 10, 2010 #1
    Some imagination is required, so let me dress up the question a bit.

    Imagine Spider-Man is on a rooftop, and he shoots a web to the top edge of a nearby building that's taller than the one he's on, in an attempt to swing toward it. Assume that the web is non-elastic and is completely adhered to the face of the building, and if he jumps off the rooftop he will naturally move in a perfect pendulum motion (ignore air resistance) A. His goal, however, is to move toward his target building following a path that is perfectly horizontal to the ground (ignore his reason) B. Keep in mind that he is able to produce a force that draws the web back into his hand, thus giving him the ability to pull himself upward if the end of the web is already attached to a surface.

    http://img706.imageshack.us/img706/5889/physics.jpg" [Broken]

    So my question is, how much force will he need to produce from his web-hand in order to maintain the perfect horizontal path? I posit that the force will need to be dynamic/changing at each point of the fall, but can the amount of force necessary to maintain the horizontal path be elegantly described in an equation that accounts for all the variables (like building distance, time taken to travel, etc)? I gave up trying once I guessed that calculus would be involved...

    If anyone is up for solving this, go for it; I'd be very interested in the solution and your thought process.


    Good to be here,
    ~ Onus
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 10, 2010 #2
    Welcome Onus.

    It won't be a pendulum motion (unless the strand is attached right ahead & he swings forthright). It will be a superposition of two pendulum motions ;a non-planar trajectory.
    If you want the strand to be of constant length , a horizontal path is impossible to maintain. If you allow elasticity, a horizontal path is possible. The calculation is not really something to funk at .
    Consider the horizontal plane in which the path lies.Assuming that the strand is taut all the times , the tension T points towards the fixed point of adherence of the strand.
    The vertical component of the tension must be mg = T sin a (m =mass of the spider & a is the angle of elevation of the fixed point from his current position).Now, the magnitude of the acceleration must be (T/m)cos a . Use polar coordinates & a simple differential equation follows.
     
  4. Mar 10, 2010 #3
    In a regular pendulum motion(the A path), the tension in the web would depend on the angle of elevation at that point in time. So you end up with T = W sin(a), where T is the tension and W is the weight of spiderman, a is angle of elevation.

    Now to travel in a straight path(B path), the total net force vertically must be equal to zero. So we consider all the forces acting vertically, W, and the new tension since spiderman is pulling downwards, the new tension is greater than before(Newton's 3rd law and all), lets call it T'.

    Now whats the value to T'? We know T' would simply be the old tension, + the added force which spiderman pulls down with, causing the web to pull back with a greater amount as well. So T' = T + F where F is the force which spiderman uses to retract his web.

    Now we have the vertical components which must equal to zero.
    So W - T' sin(a) = 0,
    => W - T sin(a) - F sin (a) = 0
    => W - W sin^2(a) - F sin(a) = 0
    => F = W(1 - sin^2(a))/sin(a)
    F = W cos^2(a)/sin(a)

    So we have the force he must pull his web in at any angle a. Now how do we find a in terms of t(time)? We have to consider the horizontal forces. Now the net horizontal force would be equal to T' cos(a) = F cos(a) + T cos(a).

    Now letting the horizontal distance of spiderman from the taller building be x, the difference in height be h, we can find cos(a) in terms of h and x.

    Using Pythagoras's Theorem, we find cos(a) = d/root(d^2+h^2), sin(a) = h/root(d^2+h^2)

    Substituting all this into the net force yields ma = W(x^3+hx)/(hx^2+h^3) which simplifies to:
    d2x/dt^2 = g(x^3+hx)/(hx^2+h^3), where g = 9.81, the acceleration due to Earth's gravity at the surface, a rather evil looking second order differential equation.
    Solve it for x in terms of t, plug it into F and you get F in terms of t, h and x.

    Oh yes another question you might want to consider is how much length of web must spiderman retract per second to obtain this motion, that is, the rate at which web is retracted.
     
  5. Mar 11, 2010 #4
    Thanks heaps, Repainted, for your easy-to-follow solution. :))
     
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