Reaction On A Inclined Plane

[SharpMan]

Hi everybody, there is a fundamental question regarding the reasoning for the observed efect in the case mentioned below. I am unable to understand How does such a thing originate (the details of the - HOW and WHAT are in the question). So please read on and be kind to explain the observed effect mentioned.

There is a metal sphere which is cut and flattened at on one side and the flat side rests on a inclined surface.

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 froem 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) a 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)


NOW THE REAL QUESTION :

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).

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 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 just solving the problem Numerically and forgetting about the actual concept behind How the reaction turned out to be greater than the applied force.

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

 

[handleman]

This thread is proof - there are some smart people who can learn the math, methods, and formulas that just are not cut out to be engineers.

-handleman, CSWP (The new, easy test)

 

[electricpete]

I don't think it's worth beating up the OP.

I think it is already well answered. But since OP maybe is looking for different slant, two thoughts:
1 - vertical forces sum to 0 and horizontal forces sum to zero, so there is no extra force. The vertial component of R2 is equal to W. We have created equal/opposite horizontal forces which are not required to counteract the weight but are required to maintain static equilibrium in this geometry. Sum of equal vertical components of W2 and R2 plus the extra horiozntal component of R2 causes R2 magnitude to exceed W2.
2 - * Parallel problem is a weight supported by two strings at angles (not vertical). Sum of tension of strings exceeds weight. It cannot be any other way since the line of the tension is not vertical, so there must be horizontal components of the tension. Since sum of vertical components of tension equal the weight, when we add in horizontal components we will certainly find the sum of tension magnitudes exceeds the weight.

The key thing is that the problem creates a constraint on direction of the reaction force. It is perhaps easier to see in the tension problem (?)

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[darai]

SharpMan:

Please post your solution to this problem so we can see how you get the numbers you post.

Maybe then I can offer a helpful comment on the questions you asked.

 

[rb1957]

i think sharpman is overthinking the solution ... why should the reactions be less than the applied load ? (consider the reactions to a torque ... the closer they are together, the bigger they are, equal and opposite and therefore in balance.

the forces (loads and reactions) need to be in balance (sum to zero). instead of thinking about what the answer should be, draw a free body and the solution will be apparent.

 

[darai]

IRstuff (Aerospace)
9 Jul 10 2:44
"If your calculated reaction exceeds the opposing force, then something must move; since nothing moves, the calculated reaction shouldn't be that value. R2 should be mg*cos(theta)"

Looking at the diagram provided, I am taking R2 to be the normal force.
If theta is the angle of the incline, that is the angle the incline makes with the horizontal, then
R2 = MG / cos theta
which is consistent with the result Sharpman obtained, since 0<= cos theta <=1