Is support force a true force ? it seems incompatibe with F=ma

In summary, the concept of "support force" is often used in physics to describe the force felt by an object that is being held stationary by another object. This force is real and can be calculated using the equation F=ma. However, in some cases, such as when discussing centrifugal force, the concept of "support force" may not be considered a true force and instead may be seen as an effect of other forces acting on the object. Ultimately, the interpretation of "support force" may vary depending on the context and perspective.
  • #1
ihaveabutt
17
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Is "support force" a "true force"? it seems incompatibe with F=ma

For example if a box is sitting on a table, not moving, it is said that the table exerts a support force on the box.

But the box and the table are both inertial right? so if F = ma (force = mass x acceleration) and acceleration is defined as the rate of change of velocity, and velocity is defined as the rate of change of position of an object, but the table is stationary, then how could there be any velocity? and if there's no velocity than how is there any acceleration? and if there's no acceleration than how can there be any force? ( m * 0 = 0 ).

Would it be proper to say that the table has no net force, if so could it still have a supportive force? is supportive force a real force in terms of F = ma or is it a not a "true force" sort of in the way centrifugal force is not a "true force"?
 
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  • #2
The box on the table feels a downward force (its weight) and an upwards force (the support from the table). The forces exactly cancel, so there in no net force and thus no net acceleration.

The table has a downward force (its weight), another downward force (the weight of the box) and an upwards force (the support from the ground) which exactly cancels the two downward forces. There is no net force on the table, so there is no acceleration.

The support force is there, it's very real, and it wants to accelerate the table. But it can't, since the weight of the table and box exactly cancel it.
 
  • #3
Centrifugal force IS a true force. Swing a weight around on a string - the force you feel is the centrifugal force, and it is the reaction to the centripetal force you apply to keep the weight rotating in a circle. Swing the weight fast enough and it will pull you over - this is clearly a true force.

Put yourself in the place of the table - do you feel the weight of the box? Is this a real force?
 
  • #4
ihaveabutt said:
F = ma
"F" here refers to the net force : the vector sum of all forces acting on the object (box). The net force on the box is zero here:

F_net = F_table_support + F_box_weight = 0
 
  • #5
MrAnchovy said:
Centrifugal force IS a true force. Swing a weight around on a string - the force you feel is the centrifugal force, and it is the reaction to the centripetal force
You are referring to the Reactive centrifugal force, which is indeed an interaction (real) force. The OP probably refers to the inertial (fictitious) Centrifugal force, in rotating reference_frames.

There is some confusion of the two, and long discussion here on the forum. Here are the differences between the two:
http://en.wikipedia.org/wiki/Reactive_centrifugal_force#Relation_to_inertial_centrifugal_force
 
  • #6
  • #7
MrAnchovy said:
there isn't a rotating reference frame here
There is no rotation at all in his example. He is not talking about some centrifugal force in his example, but just using "centrifugal force" as an example of "not a true force". So I assume he means the the "fictitious" centrifugal force, not the "real" reactive centrifugal force.
 
  • #8
A.T. said:
There is no rotation at all in his example. He is not talking about some centrifugal force in his example, but just using "centrifugal force" as an example of "not a true force". So I assume he means the the "fictitious" centrifugal force, not the "real" reactive centrifugal force.

Really? You think that while needing some reassurance in his (or her) understanding of reactive static forces he has sufficient grasp of vector mechanics to use components of its equations of motion as examples?

Or has he just been confused by a combination of being told that the force you feel when on a roundabout is not centrifugal force but centripetal force (which is of course true, but is hardly ever accompanied by an explanation of things from the roundabout's point of view), and then by the fact that he has heard the statement "centrifugal force is not a real force" from those who are talking about something completely different?
 
  • #9
Short answer: f=ma is not the only cause/effect of force.
 
  • #10
MrAnchovy said:
Centrifugal force IS a true force. Swing a weight around on a string - the force you feel is the centrifugal force, and it is the reaction to the centripetal force you apply to keep the weight rotating in a circle. Swing the weight fast enough and it will pull you over - this is clearly a true force.

Put yourself in the place of the table - do you feel the weight of the box? Is this a real force?

This disagreement happens all the time, and there seems to be no permanent resolution to it. But I have to disagree. When you are swinging a weight around on a string, the force you feel is not centrifugal force, it is tension in the string. You are pulling on the string, and the string is pulling back on you. Neither of those two forces is centrifugal force.

At the other end of the string, the string is pulling on the weight, and the weight is pulling back on the string. Neither force is "centrifugal force".

The only place where the concept of "centrifugal force" can possibly apply is right at the weight. The weight is being pulled inward by the tension force of the string, [itex]\vec{T}[/itex]. The centrifugal force is [itex]m \omega^2 \vec{r}[/itex] where [itex]\omega[/itex] is the angular velocity of the weight, and [itex]\vec{r}[/itex] is the vector from the center of rotation to the weight. These two quantities are equal and opposite:

[itex]\vec{T} = - m \omega^2 \vec{r}[/itex]

There are two different ways of interpreting the same equation:

  1. [itex]\vec{T}[/itex] is the force [itex]\vec{F}[/itex], and [itex]-\omega^2 \vec{r}[/itex] is the acceleration [itex]\vec{a}[/itex], and they are proportional by Newton's law: [itex]\vec{F} = m \vec{a}[/itex].
  2. [itex]\vec{T} + m \omega^2 \vec{r}[/itex] is the force [itex]\vec{F}[/itex], and [itex]\vec{a} = 0[/itex] is the acceleration, and they are proportional by Newton's law: [itex]\vec{F} = m \vec{a}[/itex].

So you can consider the "centrifugal force" to be a force term and put it on the same side of the equation as the tension, or you can consider it to be an acceleration term, and put it on the other side of the equation.
 
  • #11
ihaveabutt said:
For example if a box is sitting on a table, not moving, it is said that the table exerts a support force on the box.

But the box and the table are both inertial right? so if F = ma (force = mass x acceleration) and acceleration is defined as the rate of change of velocity, and velocity is defined as the rate of change of position of an object, but the table is stationary, then how could there be any velocity? and if there's no velocity than how is there any acceleration? and if there's no acceleration than how can there be any force? ( m * 0 = 0 ).

Would it be proper to say that the table has no net force, if so could it still have a supportive force? is supportive force a real force in terms of F = ma or is it a not a "true force" sort of in the way centrifugal force is not a "true force"?

I'm not exactly sure what your question is. The support force is certainly a real force. It doesn't produce any acceleration because there is a second force, gravity, that also applies to the box. Newton's law is [itex]\vec{F}_{net} = m \vec{a}[/itex], where [itex]\vec{F}_{net}[/itex] is the vector sum of all forces acting on the object. So not every force produces an acceleration, unless it is the only force acting on an object.
 
  • #12
stevendaryl said:
This disagreement happens all the time, and there seems to be no permanent resolution to it. But I have to disagree. When you are swinging a weight around on a string, the force you feel is not centrifugal force, it is tension in the string. You are pulling on the string, and the string is pulling back on you. Neither of those two forces is centrifugal force.

...

The only place where the concept of "centrifugal force" can possibly apply is right at the weight.

So if you swing a solid rod around you feel centrifugal force but if you swing a weight/string system around feeling the same force it is something different? How can you distinguish between the two different forces with your eyes closed (don't try this at home...)?

stevendaryl said:
At the other end of the string, the string is pulling on the weight

That is centripetal force.

stevendaryl said:
and the weight is pulling back on the string

That is centrifugal force - or if you insist, reactive centrifugal force although as the concept of (reactive) centrifugal force was understood hundreds of years before vector mechanics was developed and some dimwit decided to call something else centrifugal force a distinguising prefix should be used when talking about THAT force if you don't want to go around confusing people who are not speaking in the context of vector mechanics.

stevendaryl said:
This disagreement happens all the time, and there seems to be no permanent resolution to it.

Here's a partial resolution that applies in this situation: if someone is describing a system using an inertial frame of reference it is reasonable to assume that when they talk about "centrifugal force" they don't mean something that only exists in a rotating frame of reference.

stevendaryl said:
So you can consider the "centrifugal force" to be a force term and put it on the same side of the equation as the tension, or you can consider it to be an acceleration term, and put it on the other side of the equation.

Given the word "force" in "centrifugal force" and that there is no net acceleration arising in part from this force (it must be resisted by some equal and opposite force holding the centre of rotation in place otherwise there would be no rotation) I think it would be doubly peverse to consider it as an acceleration term. Do you consider centripetal force to be an acceleration term?
 
  • #13
MrAnchovy said:
That is centrifugal force - or if you insist, reactive centrifugal force although as the concept of (reactive) centrifugal force was understood hundreds of years before vector mechanics was developed and some dimwit decided to call something else centrifugal force a distinguising prefix should be used when talking about THAT force if you don't want to go around confusing people who are not speaking in the context of vector mechanics.
Well, those "dimwits" have been running the show for many decades now, at least in physics. In modern physics usage, the term "centrifugal force" (without any modifier) means the inertial (non-interaction) "force" that one adds when viewing things from a rotating reference frame. Better get used to it!

If you want to use the old-fashioned concept of centrifugal force, you'd better specify it as reactive centrifugal force. (But why even give it a special name?)
 
  • #14
To clarify, I was talking about inertial centrifugal force obviously.

Back to the topic...

Assuming the supportive force is true (whether or not it is?) then what is the net force acting on the two items (the box and the table)?

An above post says it is zero net force. That's what I assumed, but I was told elsewhere this is wrong so is there any basis for it being wrong?

And in regards to it being a true force, what exactly is the conditions that qualify a force to be considered "true"? and which of these does it or doesn't it satisfy?
 
  • #15
If the net force is not zero, then something will be accelerating. Like you said, F=ma.

It's a very easy experiment to perform. Look at the table...is anything accelerating? If not, then net force is zero :)
 
  • #16
ihaveabutt said:
Assuming the supportive force is true (whether or not it is?) then what is the net force acting on the two items (the box and the table)?
Since they aren't accelerating, the net force on each must be zero.

An above post says it is zero net force. That's what I assumed, but I was told elsewhere this is wrong so is there any basis for it being wrong?
Who said it was wrong? (Were they nitpicking about the fact that since the Earth is rotating, there is a centripetal acceleration? :rolleyes:)

And in regards to it being a true force, what exactly is the conditions that qualify a force to be considered "true"? and which of these does it or doesn't it satisfy?
Well a "real" force can be traced to some basic physical interaction, such as an electromagnetic force or a gravitational force (in Newtonian physics). In other words, real forces have actors. The normal force of the table pushing up on the box is a real force; the table is the "actor" exerting the force. "Real" forces appear in every reference frame.

Inertial forces have no "actor" since they are an artifact of viewing things from an accelerating reference frame.
 
  • #17
It seems to me that confusion is right at the beginning.
In the equation F = ma, F is the RESULTANT (net) force. It is not any particular single force.
You need to grasp this to have any chance of understanding what really is basic physics.
Someone else has pointed out that F is the net force.
 
  • #18
Doc Al said:
Well, those "dimwits" have been running the show for many decades now, at least in physics. In modern physics usage, the term "centrifugal force" (without any modifier) means the inertial (non-interaction) "force" that one adds when viewing things from a rotating reference frame. Better get used to it!

If you want to use the old-fashioned concept of centrifugal force, you'd better specify it as reactive centrifugal force. (But why even give it a special name?)

Well centuries in fact! I totally accept what you say, but I still maintain that it is not helpful to introduce the concept of inertial centrifugal force before one has a firm grasp of Newtonian mechanics in general and rotational motion in particular. It certainly confused the hell out of me until I had the maths to cope with it, and it seems to be confusing another poster on this thread who seems to know the maths but has lost the concept of reactive centrifugal force altogether!

Sorry for the hijack ihaveabutt.
 
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  • #19
ihaveabutt said:
what is the net force acting on the two items (the box and the table)?An above post says it is zero net force.
The net force on the box is zero.
The net force on the table is zero.
The net force on the table+box is zero.

See post #2
 
  • #20
MrAnchovy said:
So if you swing a solid rod around you feel centrifugal force

No, you never feel centrifugal force under any circumstances. It's not a real force. If you spin around a rod, then you feel tension forces in the rod.

Given the word "force" in "centrifugal force"

That name is a misnomer, and my opinion.

and that there is no net acceleration arising in part from this force (it must be resisted by some equal and opposite force holding the centre of rotation in place otherwise there would be no rotation) I think it would be doubly peverse to consider it as an acceleration term.

The accounting of forces and accelerations for a weight spinning at the end of a string is:
  1. The weight is acted upon by an unbalanced force [itex]\vec{F} = \vec{T_1}[/itex] due to the string.
  2. The acceleration of the weight is radially inward: [itex]\vec{a} = - \omega^2 \vec{r}[/itex], where [itex]\vec{r}[/itex] is the vector from the center to the mass.
  3. By Newton's 2nd law, [itex]\vec{F} = m \vec{a}[/itex], so we have, for the weight: [itex]\vec{T_1} = - m \omega^2 \vec{r}[/itex]
  4. The string is pulled on by the weight, according to Newton's 3rd law. So the end of the string has an outward force [itex]\vec{F} = -\vec{T_1} = + m \omega^2 \vec{r}[/itex]
  5. The string also pulls on your hand with tension force [itex]\vec{T_2}[/itex]. If the mass of the string can be neglected, then this will be the same magnitude as [itex]\vec{T_1}[/itex], but in the opposite direction: [itex]\vec{T_2} = - \vec{T_1} = + m \omega^2 \vec{r}[/itex]
  6. By Newton's 3rd law, your hand exerts a force on the string that is equal and opposite to [itex]\vec{T_2}[/itex]. So it's a force [itex]-m \omega^2 \vec{r}[/itex]

So you can account for all the accelerations and forces without mentioning centrifugal force, and in my opinion, it's confusing to even bring it up. The weight is accelerating, compared with the path of a "free" test particle, and that relative acceleration is the acceleration that is important for the application of Newton's laws.

Do you consider centripetal force to be an acceleration term?

No. It's a force. In this case, it's the tension in the string.
 
  • #21
stevendaryl said:
No, you never feel centrifugal force under any circumstances. It's not a real force. If you spin around a rod, then you feel tension forces in the rod.

If you hold the rod vertically at the top do you feel tension forces in the rod, or do you feel the normal force? If you balance the rod on your hand do you feel compression forces in the rod, or do you feel the normal force?

Consider the original post, and put yourself in the position of the box. Do you feel compression forces in the table? Or do you feel the normal force? Is this a real force?
 
  • #22
MrAnchovy said:
So if you swing a solid rod around you feel centrifugal force but if you swing a weight/string system around feeling the same force it is something different? How can you distinguish between the two different forces with your eyes closed (don't try this at home...)?

If you swing a solid rod around the force you feel is also the tension in the rod. Rod tension is pulling on your hand and that's what you're feeling as "centrifugal force"; and it is acting on the rod to accelerate the rod so that it follows a circular path around your head instead of traveling in a straight line.
 
  • #23
MrAnchovy said:
If you hold the rod vertically at the top do you feel tension forces in the rod, or do you feel the normal force? If you balance the rod on your hand do you feel compression forces in the rod, or do you feel the normal force?

I'm not exactly sure I understand the question. Are you talking about the distinction between the cause of the force, and the role of the force in the specific problem?

So a box is sitting on a table has an upward force due to--I don't know what, maybe electrostatic repulsion at the molecular level. That's the cause of the force. But whatever the cause, the role of the force is to act as a "normal force" that keeps the box from falling.

Similarly, with a weight whipping around on a string, the cause of the force on the weight is string tension, while the role of that force is to act as a "centripetal force".

So are you saying that "centrifugal force" (or "reactive centrifugal force") is a role that is played by string tension, and is the force on the person's hand that pulls outward?
 
  • #24
Force is a difficult subject. Of course, in this example there are, from a naive classical mechanics point of view, two forces acting on the body, namely the gravitational force of the Earth (magnitude [itex]m g[/itex]) and an exactly opposite contact force from the table acting precisely to the opposite direction with the same magnitude.

Microscopically this contact force comes mostly from the Pauli principle, i.e., the electrons of the atoms in the table occupy the available single-particle states, and the electrons of the body cannot occupy those, because electrons are fermions.
 
  • #25
MrAnchovy said:
If you hold the rod vertically at the top do you feel tension forces in the rod, or do you feel the normal force? If you balance the rod on your hand do you feel compression forces in the rod, or do you feel the normal force?

Consider the original post, and put yourself in the position of the box. Do you feel compression forces in the table? Or do you feel the normal force? Is this a real force?

The only forces that you can "feel" are the forces "in" yuur own body, not the forces "in" the string, the table, etc!

That might seem a pedantic comment, but it is important, because of Newton's third law - action and reaction are equal and opposite.

Of course you can measure the force in the string, etc, but that is no the same as feeling it.
 
  • #26
MrAnchovy said:
If you hold the rod vertically at the top do you feel tension forces in the rod, or do you feel the normal force? If you balance the rod on your hand do you feel compression forces in the rod, or do you feel the normal force?
Normal" is just mathematician-speak for "perpendicular to the surface", so "normal force "on an object is just the name of the force, whether compressive or tensile or anything else, that happens to be acting perpendicular to the surface of the object.

If I hold the rod vertically at the top, gravity is pulling the rod down, my hand is holding it up and the rod is in tension. If I balance the rod on my hand, gravity is pulling the rod down, my hand is pushing it up, and the rod is in compression. The compressive force in the rod is pushing on my hand and that's the normal force of the rod on my hand. My hand is pushing on the rod, and that's the normal force of my hand on the rod.

Consider the original post, and put yourself in the position of the box. Do you feel compression forces in the table? Or do you feel the normal force? Is this a real force?
If I'm the box, then I feel an upwards force from the table perpendicular to the surface of the table. Because it is perpendicular to the surface, it is the normal force of the table on me. It is produced by the table resisting the (very small) compression of the tabletop from my weight. This force exactly balances the downwards force of gravity so the net force on me is zero and F=ma holds with both F and a equal to zero.

If I'm the table, then I feel a downwards force from the weight of the box acting on me. Because this force is entirely perpendicular, it is the normal force of the box on me. This force is exactly balanced by the force of the floor pushing upwards on my legs, so again the net force is zero and F=ma tells us that the table will not accelerate. The table will, however, be very slightly compressed, and it's the resistance to that compression that causes the table to push upwards against the box.
 
  • #27
Is "support force" a "true force"? it seems incompatibe with F=ma

stevendaryl said:
So are you saying that "centrifugal force" (or "reactive centrifugal force") is a role that is played by string tension, and is the force on the person's hand that pulls outward?

Yes. I am also saying that it is IMHO no less correct (and a lot more helpful) to refer to this as (reactive) centrifugal force rather than string tension than it is to refer to the force of the table on the box as normal or support force rather than intramolecular force.
 
  • #28
Is "support force" a "true force"? it seems incompatibe with F=ma

Nugatory said:
"Normal" is just mathematician-speak for "perpendicular to the surface", so "normal force "on an object is just the name of the force, whether compressive or tensile or anything else, that happens to be acting perpendicular to the surface of the object.

Precisely. And "centrifugal" is just (Newtonian/elementary) physicist speak for "away from the centre" so "centrifugal force" on an object is just the name of the force, whether compressive or tensile or anything else, that happens to be acting away from the centre of rotation.
 
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  • #29
It helps me to think of the upward force of the Earth or an item as like a spring. gravity pulls the item downward on the spring until there is an equal upward force from he "spring" of the table or ground.
 
  • #30
MrAnchovy said:
Precisely. And "centrifugal" is just (Newtonian/elementary) physicist speak for "away from the centre" so "normal force" on an object is just the name of the force, whether compressive or tensile or anything else, that happens to be acting away from the centre of rotation.
The centrifugal "force" is a coordinate artifact. The normal force is the result of an actual physical interaction between two massive bodies (the cause is essentially Pauli exclusion principle). You can't just equate the two physically - they are quantitatively different. Things like the normal force and tension in a string attached to a point mass when the string is whirled around are examples of constraint forces. In lagrangian mechanics they come out of lagrange's equations via lagrange multipliers when you have (at the least holonomic) constraints for your system; physically they are forces that keep the object constrained to a certain path. The centrifugal "force" term on the other hand will come out when you transform from an inertial frame to a rotating one due to the derivatives of the basis vectors (they will be whirling around so they won't be constant vectors).
 
  • #31
MrAnchovy said:
Yes. I am also saying that it is IMHO no less correct (and a lot more helpful) to refer to this as (reactive) centrifugal force rather than string tension than it is to refer to the force of the table on the box as normal or support force rather than intramolecular force.

Maybe. But it seems to me that the benefit of talking in terms of "normal force" is that it hides details that are (A) irrelevant and (B) too complicated to get into. We're abstracting away from the details of how normal forces arise, and just using the property that its a contact force and is directed perpendicular to the surface of an object. Similarly, "string tension" hides the details of intermolecular forces, keeping the important facts that its direction is tangent to the curve made by the string. In contrast, "reactive centrifugal force" doesn't seem to be making the analysis any simpler, it's just giving a name to one of the several forces involved in the problem.

But it's often a matter of opinion and taste whether it is worth-while to have special names for things.
 
  • #32
WannabeNewton said:
The centrifugal "force" is a coordinate artifact

Not the way he's using it. Actually, that's another reason I don't like his terminology, is that it's hard to keep straight whether he's talking about the fictitious force, "centrifugal force", or the real force, "reactive centrifugal force".

The distinction, as I understand it, in the case of a mass swirling on a string, is that "centrifugal force" is a fictitious force, and acts on the mass, while "reactive centrifugal force" is a real force, and acts on the string. They are numerically equal, but they act on different objects.
 
  • #33
FWIW I spend a lot of my working life on the dynamics of rotating machines. I can live with the term "centrifugal force" though I prefer "centrifugal stress" which is both a real stress and self explanatory - i.e. the stress fiield caused by the rotation of the system.

On the other hand I have never (in 30 years in industry) heard anybody use the term "reactive centrifugal force", until a few recent threads in PF - and I don't see any merit in it.
 
  • #34
AlephZero said:
On the other hand I have never (in 30 years in industry) heard anybody use the term "reactive centrifugal force", until a few recent threads in PF - and I don't see any merit in it.
I've never heard the term outside of PF either (except for a wiki page) and I also see no merit in it.
 
  • #35
stevendaryl said:
The distinction, as I understand it, in the case of a mass swirling on a string, is that "centrifugal force" is a fictitious force, and acts on the mass, while "reactive centrifugal force" is a real force, and acts on the string. They are numerically equal, but they act on different objects.
Hmm, can't say I've ever heard of the "reactive centrifugal force". I'll read up on it thanks.

EDIT: I tried looking it up and I just found other forum links, a wiki link, and a facebook link (lol). Is there like some textbook this is discussed in? I couldn't find anything in Kleppner about it, which is the book I usually turn to, nor could I find anything in Taylor.
 
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<h2>1. What is a support force?</h2><p>A support force is a force exerted by an object to prevent another object from falling or sinking. It is also known as a normal force or contact force.</p><h2>2. Is support force a true force?</h2><p>Yes, support force is a true force. It may not be as obvious as other forces like gravity or friction, but it is a real force that can be measured and has an impact on the motion of objects.</p><h2>3. How is support force related to F=ma?</h2><p>Support force is related to F=ma through Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In the case of support force, it is the force that is preventing an object from accelerating due to the force of gravity.</p><h2>4. Why does support force seem incompatible with F=ma?</h2><p>Support force may seem incompatible with F=ma because it is often not explicitly included in the equation. However, it is still a part of the overall force acting on an object and is accounted for in the acceleration of the object.</p><h2>5. Can support force be negative?</h2><p>Yes, support force can be negative. This occurs when the support force is acting in the opposite direction of the object's motion, such as in the case of a car accelerating downhill. In this case, the support force would be negative as it is acting in the opposite direction of the car's acceleration.</p>

1. What is a support force?

A support force is a force exerted by an object to prevent another object from falling or sinking. It is also known as a normal force or contact force.

2. Is support force a true force?

Yes, support force is a true force. It may not be as obvious as other forces like gravity or friction, but it is a real force that can be measured and has an impact on the motion of objects.

3. How is support force related to F=ma?

Support force is related to F=ma through Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In the case of support force, it is the force that is preventing an object from accelerating due to the force of gravity.

4. Why does support force seem incompatible with F=ma?

Support force may seem incompatible with F=ma because it is often not explicitly included in the equation. However, it is still a part of the overall force acting on an object and is accounted for in the acceleration of the object.

5. Can support force be negative?

Yes, support force can be negative. This occurs when the support force is acting in the opposite direction of the object's motion, such as in the case of a car accelerating downhill. In this case, the support force would be negative as it is acting in the opposite direction of the car's acceleration.

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