Graphic representation of Static Friction Force vs. Angle of elevation

[lightng]

I'm having a hard time wrapping my head around how a graph depicting the relationship between angle of elevation of a surface (x axis) vs static friction (y-axis) illustrates a linear relationship analogous to y=x.

I understand the basic concept that as a plane becomes more elevated, eventually a threshold angle will be reached and then the object will slide down as this is a natural phenomena but i can't wrap my head around the physics behind it..

First of all, wouldn't the static force be decreasing, and not increasing if the angle goes up? Referring back to the formula if Fs = (mu)N (where the Normal force can be replaced by mgcostheta since it is on an incline plane) and since the angle approaches 90, mgcos(theta) should approach 0). This would make more sense to me because then that would mean the horizontal component of gravity (mgsin(theta)) would be able to overcome the static force, break inertia of the object and allow it to slide down the incline plane.

For reference, this is actually in the Berkeley Review Physics 2012 book.

It is Chapter 2 Passage 6: Threshold Angle Study Question #39. Thanks!

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[NextStepTutor_1]

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[DrknoSDN]

First of all, wouldn't the static force be decreasing, and not increasing if the angle goes up?

Answering this may clear up some confusion so here goes.

On a flat surface an object is only experiencing gravity and the normal force. (no friction)
As you increase the angle, if the object does not slide, the force of static friction is what is holding it in place.
As you increase the angle further gravity is producing more force parallel to the plane (F=mg sin(theta)), but if the object remains stationary you know the friction force is increasing to prevent it from sliding.

Friction force is always equal and opposite to force sliding down the ramp if the object is not moving. Therefore if you increase the slant, friction force must increase to continue holding the object in place.

[Edit]
Ahh i see, The maximum force of static friction does decrease. But when they say friction increases it is talking about how much force friction is actually applying opposite to the force down the ramp.

Near 90 degrees or "falling", maximum static friction approaches zero. But in general the force produced by static friction is not equal to the maximum force. The actual force is usually much less than the maximum because all it needs to do is counter the sliding force.

 

[lightng]

Ahhh that clears it up... I had the right intuition in thinking it was x component of gravity that leads an object to slide down an incline plane but I mis-associated it with mgcos(theta) which is why i couldn't reason it out. That makes a lot more sense. So correct me if i'm wrong but:

If the object is stationary, the static frictional force is exactly counterbalancing the x-component of gravity in a linear 1:1 fashion as the angle increases up to a point where mgsin(theta) is of greater magnitude, causing the object to slide down? If this explanation is correct then how would one determine that threshold point?

 

[DrknoSDN]

Ahhh that clears it up... I had the right intuition in thinking it was x component of gravity that leads an object to slide down an incline plane but I mis-associated it with mgcos(theta) which is why i couldn't reason it out. That makes a lot more sense. So correct me if i'm wrong but:

If the object is stationary, the static frictional force is exactly counterbalancing the x-component of gravity in a linear 1:1 fashion as the angle increases up to a point where mgsin(theta) is of greater magnitude, causing the object to slide down? If this explanation is correct then how would one determine that threshold point?

Yes, but it's not strictly the x component of gravity, it is the mg sin(theta) component. Friction is directly opposite to the force vector which is some angle below horizontal.

Threshold of maximum static friction would be FrictionS = (mu(Static) * Fnorm)
It's just the normal force times some friction coefficient that varies depending on what two surfaces are in contact.

 

[manohman]

Answering this may clear up some confusion so here goes.

On a flat surface an object is only experiencing gravity and the normal force. (no friction)
As you increase the angle, if the object does not slide, the force of static friction is what is holding it in place.
As you increase the angle further gravity is producing more force parallel to the plane (F=mg sin(theta)), but if the object remains stationary you know the friction force is increasing to prevent it from sliding.

Friction force is always equal and opposite to force sliding down the ramp if the object is not moving. Therefore if you increase the slant, friction force must increase to continue holding the object in place.

[Edit]
Ahh i see, The maximum force of static friction does decrease. But when they say friction increases it is talking about how much force friction is actually applying opposite to the force down the ramp.

Near 90 degrees or "falling", maximum static friction approaches zero. But in general the force produced by static friction is not equal to the maximum force. The actual force is usually much less than the maximum because all it needs to do is counter the sliding force.

So this is a subtle distinction, that makes sense, but i'm having trouble putting a number to.

So in such a situation where a block is on an incline and at rest, static friction must always be equal to and opposite the force of gravity.

So 0 = mgsintheta - Fstatic

In a block at incline angle theta, static friction would equal = mgcosTheta*Us
where Us is the coefficient of static friction.

if you increase the incline, the Y component of gravity (pointing down the incline) increases so the force of static friction must necessarily increase.

However, the maximum possible static friction decreases as the angle of incline is increased.

So if we're looking for the static friction force here, as the angle of incline changes does the coefficinet of static friction change? Because mgcostheta is decreasing. so the only change has to be the coefficient of static friction that should increase if it is to match the increased mgsintheta force as the incline angle is increased.

Then how does the maximum possible static friction decrease? Really confused here.

 

[DrknoSDN]

So this is a subtle distinction, that makes sense, but i'm having trouble putting a number to.

So in such a situation where a block is on an incline and at rest, static friction must always be equal to and opposite the force of gravity.

So 0 = mgsintheta - Fstatic

In a block at incline angle theta, static friction would equal = mgcosTheta*Us
where Us is the coefficient of static friction.

if you increase the incline, the Y component of gravity (pointing down the incline) increases so the force of static friction must necessarily increase.

However, the maximum possible static friction decreases as the angle of incline is increased.

So if we're looking for the static friction force here, as the angle of incline changes does the coefficinet of static friction change? Because mgcostheta is decreasing. so the only change has to be the coefficient of static friction that should increase if it is to match the increased mgsintheta force as the incline angle is increased.

Then how does the maximum possible static friction decrease? Really confused here.

First the coefficient is determined only by the materials that are being used and is independent of angle, gravity, etc, etc, so the coefficient is not changing.

You may be overlooking the Fs (less than or equal to) mgcosTheta*Us. Because it is less than or equal to it's possible that the maximum force can decrease AND the actual force can increase simultaneously. On a flat surface the two materials still have a Mu value and would resist sliding relative to one another, however there is no friction force being produced because of gravity or the normal force.

Increasing the angle of an incline will lower the max static friction because the normal force is diminishing. However if the object does not slide (still static), the actual force of friction is increasing to counteract the increasing vertical component of mgsin(theta).

The closer the incline gets to vertical the lower the maximum friction force. For clarity, it is when these two processes intersect that the object breaks free and begins sliding.

 

[manohman]

First the coefficient is determined only by the materials that are being used and is independent of angle, gravity, etc, etc, so the coefficient is not changing.

You may be overlooking the Fs (less than or equal to) mgcosTheta*Us. Because it is less than or equal to it's possible that the maximum force can decrease AND the actual force can increase simultaneously. On a flat surface the two materials still have a Mu value and would resist sliding relative to one another, however there is no friction force being produced because of gravity or the normal force.

Increasing the angle of an incline will lower the max static friction because the normal force is diminishing. However if the object does not slide (still static), the actual force of friction is increasing to counteract the increasing vertical component of mgsin(theta).

The closer the incline gets to vertical the lower the maximum friction force. For clarity, it is when these two processes intersect that the object breaks free and begins sliding.

I see. Then when looking at a scenario, how do you distinguish between when mgcosTheta*Us is the maximum static friction and not just the static friction to prevent sliding, for a given angle?

Granted when the angle is high enough such that the box is just about to start sliding, that is when the maximum static friction equals the static friction you can calculate (mggcosTheta*Us) but what about at other angles?

 

[DrknoSDN]

I see. Then when looking at a scenario, how do you distinguish between when mgcosTheta*Us is the maximum static friction and not just the static friction to prevent sliding, for a given angle?

Granted when the angle is high enough such that the box is just about to start sliding, that is when the maximum static friction equals the static friction you can calculate (mggcosTheta*Us) but what about at other angles?

To answer the first part you just need to know that friction only opposes motion so you almost never care about the maximum friction when something is sliding "due to gravity." The static friction is going to produce a force that exactly counters the force caused by gravity to keep it stationary.

The only time MGcos(Theta)*Us is going to be hitting a maximum is when more and more force is being applied while the angle is held constant (usually a flat surface for simplicity). Once the max friction is reached it breaks the static bonds and begins sliding based on the kinetic friction coefficient. A common example is a box on a flat table attached to a frictionless mass pulley system hanging off the table.
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Any MCAT question would not ask you to solve something that has multiple variables (such as "what angle will cause a box to break the static friction limit and begin sliding", because it's too complicated to plug and chug when both the force from gravity and friction are dependent on theta.)

To answer your second question, at any angle less than the maximum theta (box is not sliding), the force down the incline (MGsinTheta) is equal to and opposite of the friction force vector. However you would need to push the box down the incline (applying extra force) to reach the Maximum friction force (MGcosTheta*Us).

Just remember the actual friction balances the forces if the object is stationary. You can't have a friction vector that would end up producing a net acceleration.