Reaction On A Inclined Plane
Reaction On A Inclined Plane
(OP)
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.
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.





RE: Reaction On A Inclined Plane
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Reaction On A Inclined Plane
The vertical component of R2 is equal and opposite to W, because those are the only vertical forces.
The horizontal componenet of R2 = W tan theta = -R1
where theta is the angle of the slope to horizontal
W tan theta = -R1 because they are the only horizontal forces.
R2 is the vector sum of the two perpendicular forces, one of which is equal to W, so must be greater than W, but action and reaction are equal and opposite in whatever direction you choose to look at.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Reaction On A Inclined Plane
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Reaction On A Inclined Plane
R1 = R2*sin(theta) (the horizontal component of R2)
RE: Reaction On A Inclined Plane
Ted
RE: Reaction On A Inclined Plane
I would strongly suggest you get an elementary text on statics so that you can be enlightened to the wonders of vector forces and equlibrium.
Then you will discover that when you place s body in equilibrium, the vector amplitudes do not in general sum to zero-- only in the special case when all the forces are in line, not your case.
BTW, the more interesting question would include friction and
would result in an indeterminate answer.
RE: Reaction On A Inclined Plane
Ted
RE: Reaction On A Inclined Plane
RE: Reaction On A Inclined Plane
RE: Reaction On A Inclined Plane
This is exactly the logic I was refereing to and which did give me the reason as to WHY the reaction SHOULD be greater than the applied weight.
But it didnt give me the answer to HOW or WHAT CAUSES the reaction to be greater than the weight.
It is as if the reaction self adjusts itself or makes itself greater than the applied wight depending upon the angle of the surface so as to counter the weight in the vertical direction by its vertical component.
Thanks Hydtools for answering back, I second your view and this was the basic idea I had. But again as said above this logic limits us only to the WHY but not the HOW behind this situation.
[quote zekeman]I would strongly suggest you get an elementary text on statics so that you can be enlightened to the wonders of vector forces and equlibrium.[quote]
zekeman I have completed my elementary statics course already, and these things are often overlooked by many as they are only interested in the numerical values and not the actual Reasons or WHYs and HOWs behind things.
But I differ with this attitude and so I've asked this question which haunted me from long.
My question i.e. the HOW or WHAT MAKES THE REACTION GREATER is still not answered. Can you enlighten all of us by elaborating on HOW the reaction gets itself greater than he weight applied.
Can you please site an example of this it will be nice to learn and correct my concepts if they are wrong somewhere.
Guys, said agian, my question is still unsolved as to HOW does the reaction become greater than the applied wight. WHERE does the extra force required ( to make the vertical component of R2 equal to W) come from.
Please guide mme through this with a Reason for the observed effect of the reaction.
RE: Reaction On A Inclined Plane
RE: Reaction On A Inclined Plane
RE: Reaction On A Inclined Plane
yes ... it is the vertical component of the reaction that is reacting the load. because the reaction is inclined it is bigger than the applied load ... no mystery, no rocket science. this creates a lateral compoent (of R2) that is balanced by R1.
if R2 was vertical and under the load (same line of action) then R1 would be 0 and R2 = Wt.
RE: Reaction On A Inclined Plane
RE: Reaction On A Inclined Plane
Then you will discover that when you place s body in equilibrium, the vector amplitudes do not in general sum to zero-- only in the special case when all the forces are in line, not your case
Can you please site an example of this it will be nice to learn and correct my concepts if they are wrong somewhere."
Let me try.
Suppose you wish to hang a weight in the center of a tight horizontal string supported at its ends.
You will get a slight sag in the center and the the forces that keep the weigh in equilibrium are the 2 string tensions and gravity pulling down the mass.
Now the 2 tensile vector forces have to deliver a vertical component of force = to the weight; the symmetric angles the string makes to the horizontal is @. Let's say the angle is 0.1 radian for example so the vertical statement of equilibrium is
2Tsin@=weight=W
sin@=.1
So,
T=W/.2=5*W
So the tension in each segment of the string is 5 times the weight.
Pretty elementary statics , and clear ( at least to me).
Note,if the weight were not centered on the string you would get unequal angles and tensions in the segments.
RE: Reaction On A Inclined Plane
True.
But the horizontal components are equal and opposite. Thus the vector sum of horizontal components is zero.
The vertical components sum to equal and opposite W. Thus for the system the total sum of forces is zero.
If the sum of forces on a system is not zero you have acceleration.
RE: Reaction On A Inclined Plane
In terms of work and energy, a small force moves a large distance to move a larger force a smaller distance. Conservation of energy requires that the work of each force are equal.
Ted
RE: Reaction On A Inclined Plane
The example you gave in the OP is an example of it. The string with a force applied to the middle is another example, perhaps intuitivly easier to grasp (although I'm not quite sure why it is).
I think the thing you are missing is that it is the vector sum of the forces that are equal for a body in equilibrium. The horizontal forces cancel out, leaving a vertical component of R2 that is exactly equal to W.
If you now choose to combine the vertical component of R2 with its horizontal component, the resultant must necessarily be greater than W.
The answers are in the thread; you just need to change your mind set a little to see it. Maybe changing the question would help. Would you expect the resultant force R2 to be equal to W, and if it was how would it be possible for the thing to be in equilibrium?
Don't let the responses to your question put you off asking about the basics though. Thinking about these things at that level really is the only way to understand them properly, as opposed to just applying a learned procedure.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Reaction On A Inclined Plane
"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."
This is just plain wrong, and is not intuitive to me. A lever is another mechanism that can result in reaction forces much higher than the applied force. There is no such thing as a "Law of Conservation of Forces" that decrees that reaction force magnitudes can't exceed applied force magnitudes. Why wouldn't intuition instead tell you that the applied force can't exceed the reaction? 'Applied force' is only your perception of what is being applied. How is the weight of the sphere any more of an 'applied' force than the two reactions being applied to it??
Basically if you want an answer to the question 'How can a reaction force be higher than the applied force?' I would say 'Because we live in a universe governed by physical laws that allow it to be so'.
RE: Reaction On A Inclined Plane
-handleman, CSWP (The new, easy test)
RE: Reaction On A Inclined Plane
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 (?)
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
RE: Reaction On A Inclined Plane
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.
RE: Reaction On A Inclined Plane
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.
RE: Reaction On A Inclined Plane
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