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.

 

[IRstuff]

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)

TTFN

FAQ731-376

 

[IDS]

Assuming we are treating both the inclined and vertical planes as frictionless:

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/

 

[IDS]

I would add that I prefer to look at a force as being an interaction between two bodies acting along a line, rather than an action by one body on another, acting in a particular direction.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

R2 = Wt/cos(theta) (the vertical component of R2 is Wt)
R1 = R2*sin(theta) (the horizontal component of R2)

 

[hydtools]

R2 is greater than the weight because it is the reaction to both the weight and the force of the vertical against the sphere. As the inclined angle becomes smaller towards horizontal the reaction R2 becomes smaller tending toward equaling W when the angle reaches 0 deg. As the incline angle becomes smaller the reaction of the vertical wall becomes smaller toward equalling 0 when the angle equals 0 deg.

Ted

 

[zekeman]

Sharpman,

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.

 

[hydtools]

zekeman, including friction forces would not result in an indeterminate solution. The surface normal and friction forces can be resolved into single resultant reaction forces. The resulting three-force system on the sphere can then be solved.

Ted

 

[rb1957]

good point, the friction forces are a fraction of the normal force, so there are still only 2 unknowns

 

[zekeman]

The reason that it is indeterminate is that the friction coefficient is anywhere from zero to the static value, depending on how it is loaded.

 

[SharpMan]

Thanks IDS,

R2 is the vector sum of the two perpendicular forces, one of which is equal to W, so must be greater than W

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.

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.

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.

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.


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.

 

[SnTMan]

something like this:

 

[SnTMan]

Or maybe more like this

 

[rb1957]

"But it didnt give me the answer to HOW or WHAT CAUSES the reaction to be greater than the weight."

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.

 

[MintJulep]

The vector sum of R1 + R2 = W

 

[zekeman]

" Quote (zekeman):
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.

 

[MintJulep]

So the tension in each segment of the string is 5 times the weight.[/quote}

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.

 

[hydtools]

This question illustrates the mechanical advantage of the wedge. A small wedge angle allows a small force create or require a large reaction to prevent motion.

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

 

[IDS]

Can you please site an example of this it will be nice to learn and correct my concepts if they are wrong somewhere.

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.

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.

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/

 

[FullMetalBracket]

I believe hydtools has properly answered the question. Your main problem is the following statement:

"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'.