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Larger Reaction Than The Applied Force

  1. Jul 9, 2010 #1
    There is a metal sphere which is cut and flattened at on one side and the flat side rests on an inclined surface.

    The figure is attatched please refer it.

    The weight of the sphere is W = 20*9.81 N .

    The sphere will tend to slide on the inclined surface as perceived. But the vetical wall on the left hand side prevents it from doing so and the sphere is in static equilibrium.

    The reactions generated at both the supports are R1 ( at vertical left wall) and R2 (at the inclined surface acting Normal to the surface) as shown.

    The working out of the Equilibrium Equations the reaction R2 at comes out to be 305.23 N

    ( We do not discuss about reaction R1 here as it is not of our interest)


    The Reaction to any force is generally (to our general perception or natural instinct) equal to the force or a fraction of the force if the force is acting at an angle to the support ( the sine or cosine term like Wsinθ or Wcosθ)

    But we never think of or perceive the reaction to EXCEED the Acting Force.

    However in the example above the Reaction R2 (R2 = 305.23 N) is much more than the ONLY ACTING FORCE i.e. THE WEIGHT of the sphere (W = 20*9.81 = 196.2 N)

    I've been wondering from WHERE does this EXTRA FORCE come from or WHAT GENERATES IT if the ONLY ACTING ( AVAILABLE SOURCE ) OF FORCE is THE WEIGHT i.e. ONLY 196.2 N.

    What happens and how does the reaction turn out to be greater than the apllied force.

    I couldnt figure out the actual REASON or PHYSICS behind this happening. Neither could I simply sit down satisfied with the Numerically correct answer and forgetting about the actual concept behind How the reaction turned out to be greater than the applied force.

    And What generated or developed this reaction which Exceeds the Force Applied.

    So can anyone please explain this phenomena as to How the reaction becomes greater than the applied force.

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  3. Jul 9, 2010 #2


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    A "reaction force" is an equal and opposite force an object applies back against something applying a force to it. So the force applied at an angle by the wall is not a reaction force to the gravity. This is basically just a leverage issue.
  4. Jul 9, 2010 #3

    Then what is it? The static equilibrium has reached (the spherical object doesnt slide down along the inclined surface anymore) so there has to be Reactions which to best of my knowledge are responses to the forces or their components (in case of inclinations) acting on the surfaces/supports.

    Can you explain this. How has the leverage originated?
    Last edited by a moderator: Jul 9, 2010
  5. Jul 9, 2010 #4


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    Well, first off, by definition a "reaction force" is equal and opposite of an applied force. So since the force angled off to the side is neither equal nor opposite, it isn't the reaction force.

    But what you can do is take the appropriate vector components of the various forces and add them together to obtain the reaction forces. So using trigonometry, if you calculate the vertical and horizontal components of R2, you'll find it contains a horizontal component which opposes R1 and a vertical component that opposes W and all of the forces sum to zero.

    This is the essence of the subject "statics" - a sophomore level engineering course.
  6. Jul 9, 2010 #5


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    Also, the relationship to leverage is that gears, pulleys, levers, and angular application of forces are all different ways of achieving a mechanical advantage.
  7. Jul 9, 2010 #6
    I do understand Leverage and Mechanical Advantage in terms of pulleys, levers (moments causing it), winches et al.

    But the question here is not limited to sophomore statics. I do understand that reactions are equal and opposite to the acting force. In general we consider the reactions in the component directions as well.

    But in this case (or any inclined surace/support or force acting at an angle) we usually resolve it in the appropriate directions and then consider each of the two components to act vertically and horizontally (considering X-Y axis as principle ) or resolve the force into components along the normal and horizontal to the incllined surface (as we would do in this case if the sphere had only to slide down the inclined surface - Wcos40 would be acting normal to the surface and Wsin40 would be acting parallel to the surface providing for the downward slip/sliding of the sphere along the inclined surface - had there been no vetical wall on the left side)

    Yet my question remains un answered as to HOW the so called reaction R2 , ALL ON ITS OWN INCREASES ITS VALUE ( GREATER THAN THE APPLIED WEIGHT) to maintain the Equilibrium.

    Where does this extra force come from.
  8. Jul 10, 2010 #7


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    StartlerBoy: There is a mistake in your current answer for R2. R2 should instead be 256.12 N. Try it again.

    Secondly, any (passive) force applied by a support is a reaction force, regardless of the coordinate system you use to express it. R2 is a reaction force. R1 is a reaction force.

    Moving on to your main question, the extra force you are asking about comes from the wall. Your surfaces are frictionless. If the wall were not there, the sphere would slide down the inclined surface. The wall stops the sphere from sliding. The reaction force applied by the wall goes through the sphere, and into the inclined surface. This is extra force on the inclined surface.

    In summary, if you try to hold up a mass by pushing at an inefficient angle, you must apply more force. Try it. Hold up a mass by pushing straight upward. Now try to hold up the same mass by pushing at an angle, instead of upward.
  9. Jul 10, 2010 #8


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    Again, you're interpreting it wrong. As I already said, r2 is not the reaction force as it is neither equal in magnitude nor opposite in direction. The reaction force is only the vertical component of r2 and it is equal to and opposite of W.

    However, the reason r2 can be greater in magnitude than W is, as I also already said: Leverage!

    Try this link - maybe it'll help clarify for you: http://physics.about.com/od/physicsintherealworld/p/simplemachines.htm
    Or this: http://en.wikipedia.org/wiki/Wedge_(mechanical_device [Broken])

    Yes, there is more than one way to analyze the situation, but they all achieve the same answer. If you want to orient your axes along the plane, fine: then you also need to orient the force of gravity along the angled plane. You'll have one component parallel and one component perpendicular and the forces all still sum to zero.
    Last edited by a moderator: May 4, 2017
  10. Jul 11, 2010 #9
    Thanks Guys for your responses.
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