# Physics Question and point about angle's and friction

Discussion in 'MCAT: Medical College Admissions Test' started by christian15213, May 1, 2007.

1. ### christian15213 Membership Revoked Removed

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when doing the angled plane you must first take two things into consideration. Is your object moving or not? If it is not moving then this is everything you need to know in regards to forces...

1. with NO friction the only force acting on a non-moving object (box per se) is going to be Fn... which in this case is the Fn from the Y axis and there for mgcos0... If the box is moving and still NOT considering friction then you must take Fn in conjunction with gravity which is not moving the object along the X axis. This is going to be mgsin0...

2. With friction the same rules apply but no we must consider how we are looking at friction. If the box is not moving, then static friction is going to be the reason why. But in this case... static friction is actiing on the box in the X axis opposing gravity equally to keep the object stationary... thus the equation used is going to be umgsin0... Now, when the box is moving and this is where I might need some help understanding the how to apply all the forces... the friction in this case is Kinetic... but it is constently acting from the Fn y axis while sliding down... thus the equation is umgcos0 but is drawn as a force that is opposing the forward motion. and there for has to be subtracted from the orginal sliding force from equation set 1 above mgsin0... ??? is this an accurate explanation. Am I thinking of this correctly...

one other thing yea I know there could be tension that I am fine with... but in regards to the stationary box again... the the equation would be mgsin0 - umgsin0 = 0 which means no movement...

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anyone want a crack at this?

4. ### sehnsucht 2+ Year Member

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here is what I think:

if the box is static, the equation should be equal to 0 to reflect equilibrium in the Y and X directions. as such, the vertical equation at eq would be N-mgcos[SIZE=-1]&#952; = 0. in the horizontal, the equation at eq would be F(friction) - mgsin[/SIZE][SIZE=-1]&#952; = 0. if the box is moving the force acting on the box would be the net force of the Fn minus the kinetic force of friction. the equation for that would be: Fn - Fk = Fnet = 0

there would be no tension unless the box was physically attached to anything else. consider only the forces that are physically touching the box to get a better idea of what forces apply to the problem. in your case, they are all "contact forces" and they include the Normal, mgsin[/SIZE][SIZE=-1]&#952;, Friction (F=uN) and mgcos[/SIZE][SIZE=-1]&#952;. [/SIZE][SIZE=-1]the only force keeping the box from sliding full bore down the incline would be the friction, acting in the opposite direction to your box movement, not tension.
[/SIZE]

5. ### trustforward 7+ Year Member

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i wish i had a better grasp on phys to actually explain in my own words,but I'm not going to kid myself. anyway, try this out to see if it will set you straight:
http://forums.studentdoctor.net/showpost.php?p=3587973&postcount=24

6. ### trustforward 7+ Year Member

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i wish i had a better grasp on phys to actually explain in my own words,but I'm not going to kid myself. anyway, try this out to see if it will set you straight:
http://forums.studentdoctor.net/showpost.php?p=3587973&postcount=24

7. ### HippocratesX Member 7+ Year Member

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I'm not sure what your question is, but on an incline plane problem...say a stationary box, the normal force is not the only force acting on the box. There is also a force of Mgsin(angle) that is parallel to the incline plane. It doesnt matter if there is fiction or not, this is still a force acting on the box.

8. ### cheezer Guest

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Box on an angled plane

Step 1. Draw a coordinate system.
x-axis parallel to the surface of the plane. y-axis perpendicular to the surface of the plane. label the postive and negative axes. +y is up along the direction of the y axis. -y is down along the direction of the y axis. +x is right along the direction of the x axis and -x is left along the direction of the x axis.

Case 1: Frictionless Surface, Box not moving.
Fx=0
Fy=0
This should be automatic.
You have a normal force perpendicular to the surface along the +y-axis
+Fn
You have a component of mg along the -y-axis. This is mgsin(theta) or mgcos(theta) depending on what angle they give you, let's call it mgsin(theta) for now.
-mgsin(theta)
Sum the forces. Fy=[+Fn]+[-mgsin(theta)]=0
Now look at Fx
Since there is no friction, there is no force along the postive +x axis.
0
However, there is a component of mg along the -x-axis. let's call it mgcos(theta) for now.
-mgcos(theta)
Sum the forces. Fx=[+0]+[-mgcos(theta)]=0

Case 2: Frictionless surface, Box is moving.
The only difference between Case 1
Fx=[+0]+[-mgcos(theta)]=ma <--since there is now acceleration
Fy=[+Fn]+[-mgsin(theta)]=ma=0 <--since there is no acceleration along the y component since your coordinate system has the y-axis perpendicular to the surface of the plane.

Case 3:Surface with friction, Box is not moving
Case 1's formula's, except now you add static friction force along the +x axis

Fx=[+UsN]+[-mgcos(theta)]=0 <--N=Fn=mgsin(theta)
Fy=[+Fn]+[-mgsin(theta)]=0 <--this remains the same

Case 4: Surface with friction, Box is moving
Case 3's formula's, except static friction force along the +x axis is now kinetic friction force along the +x axis.

Fx=[+UkN]+[-mgcos(theta)]=ma <--N=Fn=mgsin(theta)
Fy=[+Fn]+[-mgsin(theta)]=0 <--this remains the same

To answer your question your problem lies in how you draw your coordinate system. Draw it right and it eliminates the component problems. Btw I'm assuming the plane is a right triangle, the surface is the hypotenuse and the box is sliding down the hypotenuse.
oh yeah one more thing. Tension only occurs when the box is held on the plane by a rope. In this case you should just add a +T to the Fx summation. Remember that all of the above is simply a free body diagram for a single mass. If there is a system of masses, seperate them into individual free body diagrams and write out all their Fx and Fy summations using the same technique as I described above.

...Did that answer your question? I hope so, it took me 10 minutes to write that ****.

9. ### cheezer Guest

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bump

10. ### christian15213 Membership Revoked Removed

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I think you guys have it amazingly wrong...

Both of your cases are wrong I believe...

when the box is not moving with no static forces enacted... that is pure Fn = mgcos0

when the box is moving with no friction that is mgsin0... I described the reasons why...

it doesn't matter what the angle is including zero angle which means horizontal or 90 which means vertical drop...

am I missing something? cause I was sure about those parts...

11. ### CATallergy 2+ Year Member

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OP, why are you posting this? if you actually read all the crap on this page and think it is "amazingly wrong", then you obvious are confident enough about this material not to need help. are you [god knows why] trying to establish some general approach to incline problems?

1) draw a diagram

2) break all forces into matching perpendicular vectors (however seems easiest)

12. ### sehnsucht 2+ Year Member

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i don't quite follow where this post is going and its original intent so humor me - you had a question, we sought to help you out. the forces that break down are quite straight forward and the suggestions people put up coincide with solutions I have seen in EK, TPR and Nova Physics. different approaches to the logic still check out which works out fine, i think. The forces acting can only be in the X, Y and due to friction if you draw out the different scenarios. so, the million dollar question is: why are you asking the question if you know we are wrong?

13. ### cheezer Guest

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You mean you don't draw a coordinate system on a free body diagram, break it down into Fx and Fy components and solve for the unknowns?

Then it's you who doesn't understand mechanics.

Again, besides the above system the only differences in intepretation is how you draw your coordinate system and what angles you can work with.

Fn can =mgcos(theta) or it can =mgsin(theta) depending on where you take your angle and if you draw your coordinate system in a way where the y-axis is perpendicular to the surface the box is sliding down. And when you're taking the component of a force, the angle completely matters.

It is not set in stone...it is trigonometry.

So you say Fn=mgcos(theta) (I seriously hope that is not the number 0 you have down there), then the x-component of your force is mgsin(theta). Fn does not magically change into mgsin(theta) just because the box is moving.

If you choose the right coordinate system (ie y-axis is perpendicular to the surface the box is sliding down), the Fy forces stay the same whether the box is moving or not.

I am not amazingly wrong, I am amazingly right, as are the other posters who tried to help you. I only elaborated on what they mentioned. Trust me when I say that classical mechanics is the easiest thing ever and if you bothered to try and understand my previous post then you'd know that.

Seriously, if you can't solve a problem like this, you're f*cked and if you think what we've suggested is wrong, then you're f*cked.

14. ### cheezer Guest

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Ugh, I'll try to answer this more specifically

You:when the box is not moving with no static forces enacted... that is pure Fn = mgcos0

This is partly true if the following conditions are met. btw im assuming you mean the static friction force.
You draw the coordinate system like I described above, thus Fn is an Fy force.
Fn=mgcos(theta) if the angle you are taking is the angle between the perpendicular of the surface and the direction of the mg.
Fn=mgcos(theta) is not the only force acting on it. After all you are splitting the weight of the box into components. So what about mgsin(theta)? That is the force acting
on the box along the surface of the angled plane. Just because something isn't moving doesn't mean that this force doesn't exist, it just means the net force is 0.

You:when the box is moving with no friction that is mgsin0... I described the reasons why...
When the box is sliding down the frictionless plane, yes, the force that is pushing it along the plane is mgsin(theta). Your Fn, however, stays the same i.e. mgcos(theta)

You:it doesn't matter what the angle is including zero angle which means horizontal or 90 which means vertical drop...
this is all a matter of reference, i.e. again, how you draw the coordinate system.

As for part two:

You: If the box is not moving, then static friction is going to be the reason why. But in this case... static friction is actiing on the box in the X axis opposing gravity equally to keep the object stationary...

Yes, if the box is not moving, that means the static friction force is equal to the force pushing on the box along the surface of the plane.

You: thus the equation used is going to be umgsin0...

No. it is fs=Fx in other words, according to the conventions in this post, [Us*mgcos(theta)]=mgsin(theta) where mgcos(theta) is your normal force

You: Now, when the box is moving and this is where I might need some help understanding the how to apply all the forces... the friction in this case is Kinetic... but it is constently acting from the Fn y axis while sliding down... thus the equation is umgcos0 but is drawn as a force that is opposing the forward motion. and there for has to be subtracted from the orginal sliding force from equation set 1 above mgsin0... ??? is this an accurate explanation. Am I thinking of this correctly...

What? Now you have the case where fk-Fx=ma
[Uk*mgcos(theta)]-mgsin(theta) = ma

You: one other thing yea I know there could be tension that I am fine with...

Only if it's attached to a rope

You: but in regards to the stationary box again... the the equation would be mgsin0 - umgsin0 = 0 which means no movement...

NO. Written purely for the x-axis, it would be Fs-Fx ---> [Us*N]-mgsin(theta) ---> [Us*mgcos(theta)]-mgsin(theta)
You can also write it Fx-Fs if you decided to call the direction the box slides down the postitive x axis and it's opposing frictional force the negative x axis. It really doesn't matter, the answer will adjust for it.

That's it, I'm done.

15. ### christian15213 Membership Revoked Removed

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Ok, but what I am saying is true... I am trying to confirm the, OK this is correct... that is all. But I am reading what you guys are saying and I am thinking it is wrong.

Does anyone else agree that might have a handle on the subject... .

16. ### christian15213 Membership Revoked Removed

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Ok, but what I am saying is true... I am trying to confirm the, OK this is correct... that is all. But I am reading what you guys are saying and I am thinking it is wrong.

Does anyone else agree that might have a handle on the subject... .

17. ### christian15213 Membership Revoked Removed

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ok, now I think you might have it... I will read it over...

18. ### CATallergy 2+ Year Member

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It's hard to tell exactly what you're thinking, because looking at someone else's thought process can be confusing. But it looks like your off a little.

The force of friction (&#956; mg cos &#952 is a force acting in the direction opposing the motion, not gravity. Thus, when the block is not moving, &#956; mg cos &#952; = mg sin &#952;.

You are right that it is actually gravity causing the force mg sin &#952;, but this is only part of mg, and the rest (mg cos &#952 is being countered by the normal force of the plane pushing back against the block (preventing the block from sinking into the plane).

19. ### christian15213 Membership Revoked Removed

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Thanks for understanding me... I was thinking I was going crazy till I read some of those other post's... but hey, we're here to learn...

Ok, with this statement you make...

****The force of friction (&#956; mg cos &#952 is a force acting in the direction opposing the motion, not gravity. Thus, when the block is not moving, &#956; mg cos &#952; = mg sin &#952;*****

The way EK explains it and the way that is making sense to me... is that Fn is u mg cos O but in the case of static (non-moving) friction the force of gravity is acting on the block... pushing it down... however, it just so happens that the static frictional force is equally action on the object in the opposite direction as if it where going back up the angled plane on the x access.. so there for it has to be mg sin O...

untill this static force is overcome the block will remain static... if it is overcome the box will begin sliding and thus you have a force pushing the box down which is gravity... now, for this we dont' use static friction we are now using kinetic and the k friction will act to "slow" the box down but it is not an opposing X axis force... but rather a constantly y acting force that is sort of going along for the ride... that is why it is mg cos O and not sin... even though when you draw the forces you still be the frictional pointing in the opposite of motion direction...

Again, I think I am right I just want a conformation...

20. ### bluesTank Zombie 5+ Year Member

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First of all, it is impossible to have a stationary box on an inclined plane on the surface of the earth if there is no friction involved, unless there is something obviously holding it back, like tension up or something. You just draw your free body diagram and find the net force on the object.

Normal force is going to cancel out with the cos component of the gravity force, and your net force will be wsin(theta) - your frictional force.

Not sure what you are asking still.

21. ### cheezer Guest

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Pretty damn clear. The motion he mentions is mg sin &#952;

If you're referring to the Normal force, it's mg cos &#952;, without the u

&#956; mg cos &#952; = static frictional force opposing the motion downward

mg sin &#952; = motion downward

force pushing the box down the ramp remains mg sin &#952;

net force =0 before the box begins moving

kinetic frictional force is less than static frictional force, that's why the box begins moving downward.

Refer to above. kinetic friction will slow the box down, just not to the point that it stops the box like static friction does. kinetic friction IS an opposing X axis force to the motion (if you consider the ramp the x axis)

Completely wrong. the only Y acting force is the normal force, which is mg cos &#952;. Think of kinetic friction as a lesser strength of static friction.

even though? the reason they're drawn that way is because they have nothing to do with the Y forces.

Not completely right. Not completely wrong.

22. ### Rofeh20 Kaplan MCAT Instructor 5+ Year Member

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I did not read through every post carefully, so my apologies if I am being repetitive. If the box is not moving before and after you change the angle of inclination, then the only force to concern yourself with is the force of static friction.

I like to think of the force of static friction as having two equations:
1) Fs = Fapplied (presuming that Fapplied < Fsmax)
2) Fsmax = (mu of static friction)(normal force) = (mu of static friction)((mg * cos(theta))

Since the box was not moving before and after the angle of inclination was increased we know that Fapplied is still less than Fsmax (if that was not true then the box would be moving).

So, Fs = Fapplied. Fapplied is the component of the force acting on the box to cause it to slide down, i.e. Fapplied = Fg * sin (theta). As you increase the angle, you are increasing the Fapplied. Since Fs = Fapplied, as you increase the angle, you are also increasing Fs.

We can compare this to a stationary skier. The magnitude of the force of static friction holding the skier in place will increase with the steepness of the slope. For example, let's say we have a 100 N skier on a slope. In all scenarios the skier is not moving before and after we increase the angle. At a slope of zero degrees, the force applied is equal to zero as the sin(0) is zero. The force of static friction must also be zero in this case (recall that frictional forces only act in opposition to other forces...if no force is acting on the skier to pull it down the slope then Fs must also be zero). At a slope of 30 degrees, the force applied is equal to 50 N as sin(30) is equal to 1/2. Since we know the skier is not moving, the force of static friction must also equal 50 N as well. So, as the angle of inclination increases, the force applied increases and therefore the force of static friction acting to oppose the force applied will also increase.

The potential confusion lies with the fact that the maximum force of static friction is ALSO changing at the same time! And it is decreasing! As noted above, Fsmax is a function of mg * cos(theta). As theta increases, cos(theta) will decrease; therefore the steeper the slope the lower the maximum amount of static friction. Once again, if you apply that to a real world situation that will make sense...as the steepness of a slope increases less applied force is necessary to get the block to move.

BTW, sticking with the skiing analogy is an easy way to remember that you are interested in the sin(theta) component of gravity. When theta = 90 degrees, that means you are on a perfect cliff. In such a situation your contact with the ground will be minimized, and mg * sin(theta) is simply equal to mg, as the sin(90) is 1. So, if you confuse sin v cos in these types of incline plane problems just think about this scenario and you'll be able to figure it out.

In general, I would recommend pausing to reflect upon the formulas rather than memorizing them. Only a minority of questions (if any) will ask you to spit back formulas; most of the questions you will encounter on the test will test your ability to understand the passage and apply what it is talking about to the basic content you know to answer the questions.

I hope that helped.