Cannot understand Normal Force, Solution to problem

  • #1
NoobeAtPhysics
75
0

Homework Statement



I am studying the solution used for the following question:

http://answers.yahoo.com/question/index?qid=20131014213456AA4eCNl

And I am unable to understand why

normal force Fn = mgcosΘ + PsinΘ

Should it not be just Fn = mgcosΘ, When drawing the FBD, the only forces that act in the Y direction are Fn and Fgcos(theta) , so Fn = -Fgcos(theta)

Homework Equations



F=MA

The Attempt at a Solution



Please see above.


Thank you,
Nobe
 

Answers and Replies

  • #2
nasu
3,957
583
How about P's component in the y direction?
 
  • #3
MGCLO
18
0
a guy already answered the question on yahoo answers!
 
  • #4
NoobeAtPhysics
75
0
Yes, I am looking at the solution but I don't seem to understand it.

nasu, if we turn out heads (my teacher told me to do this), we see that the only forces acting in the y direction are Fgsin(theta) and Fn, P is in the x direction if we turn our heads
 
  • #5
nasu
3,957
583
I have no idea what you mean by turning heads.
But the components of P are independent of what we do with our heads.
If you call "Y" the direction perpendicular to the inclined plane, then P has a component along it. P is not perpendicular to Y.

If you call "Y" the vertical direction, then indeed P has no component about this "y". But then the weight has no component about this so there is no mgcosθ and N has component Ncosθ.

So what do you call "Y" direction? Maybe you should start here to avoid confusion.
 
  • #6
MGCLO
18
0
Yes, I am looking at the solution but I don't seem to understand it.

nasu, if we turn out heads (my teacher told me to do this), we see that the only forces acting in the y direction are Fgsin(theta) and Fn, P is in the x direction if we turn our heads

Normal Force is equal to the force acting on the surface. which is in case Y-Component of G.and it must be perpendicular to the surface. so when we have an angle over here. you need to get the Y-component of the FG(Force of Gravity). which is equal to the normal force. if you don't have an angle. Fn(Normal Force) will simply equal FG. but in this case you do!
 
Last edited:
  • #7
NoobeAtPhysics
75
0
nasu,

if we rotate our heads by theta, Y direction is the direction as we see it in this angle.

MGCLO,

I understand this, but why is the P force being used for the calculation of the Fnet in the y direction?
 
  • #8
22,143
5,069

Homework Statement



I am studying the solution used for the following question:

http://answers.yahoo.com/question/index?qid=20131014213456AA4eCNl

And I am unable to understand why

normal force Fn = mgcosΘ + PsinΘ

Should it not be just Fn = mgcosΘ, When drawing the FBD, the only forces that act in the Y direction are Fn and Fgcos(theta) , so Fn = -Fgcos(theta)

Homework Equations



F=MA

The Attempt at a Solution



Please see above.


Thank you,
Nobe
The normal force is not acting in the y direction. It is acting in the direction perpendicular to the inclined plane. That is what the term normal force means. The force P has a component perpendicular to the incline (Psinθ), and the gravitational force has a component perpendicular to the incline (mgcosθ). The normal force balances these other two forces.

Chet
 
  • #9
NoobeAtPhysics
75
0
Ok, normal force is not acting in the Y direction when we look at the diagram normally. I understand.

But when we rotate our heads to see the diagram, we get that

Fnetx = P -Ff-Fg*sin(theta)

Fnety = Fn - Fgcos(theta)

I'm still confused why P is part of the normal force

This is the method i've used for a lot of problems but now i feel confused
 
Last edited:
  • #10
NoobeAtPhysics
75
0
have i been doing it wrong all this time?
 
  • #11
462chevelle
Gold Member
305
9
I don't know why you need to rotate your head. but if you look at the fbd the x axis will use cos and the y axis will use sin
 
  • #12
22,143
5,069
Ok, normal force is not acting in the Y direction when we look at the diagram normally. I understand.

But when we rotate our heads to see the diagram, we get that

Fnetx = P -Ff-Fg*sin(theta)

Fnety = Fn - Fgcos(theta)

I'm still confused why P is part of the normal force

This is the method i've used for a lot of problems but now i feel confused

OK. Let's do the force balance in the y (vertical direction) and see what we get. P has no component in the y direction, and the gravitational force in the y direction is mg. Let Fn represent the component of the force the plane exerts on the block in the direction normal to the plane, and let Ff be the component of the force the plane exerts on the block in the direction tangent to the plane. The component of Fn in the y direction is Fncosθ, and the component of Ff in the y direction is Ffsinθ, so the force balance in the y direction is as follows:

Fncosθ+Ffsinθ=mg

Lets continue with the force balance in the x direction. The gravitiational force has no component in the x direction, but the force P acts in the x direction. The component of Fn in the x direction is -Fnsinθ, and the component of Ff in the x direction is Ffcosθ. So the force balance in the x direction is as follows:

Ffcosθ+P=Fnsinθ

Now, if we multiply the y force balance by cosθ and the x force balance by sinθ and subtract them, we end up with

Fn=mgcosθ+Psinθ

So, either way you do the force balances (however you want to tilt your head), you end up with the same result.

Chet
 
  • #13
NoobeAtPhysics
75
0
using the non-tilt method, shouldn't it be:

Fncosθ+Ffsinθ - Psinθ=mg [You are considering down to be positive?]

since we are considering all the forces in one direction?? P does have a y component though to consider right?


wow................. i guess i was doing it wrong the whole time, im really confused now.
 
Last edited by a moderator:
  • #14
22,143
5,069
using the non-tilt method, shouldn't it be:

Fncosθ+Ffsinθ - Psinθ=mg [You are considering down to be positive?]

since we are considering all the forces in one direction?? P does have a y component though to consider right?

Just look at the diagram. P is horizontal (oriented purely in the x direction), and doesn't have a component in the y direction. Another way to think about it is: P and mg are perpendicular, and mg is purely in the y direction, so P must be purely in the x-direction.
 
Last edited by a moderator:
  • #15
nasu
3,957
583
From the way he wrote the components initially (and the head tilting) it seems that the OP considers the direction normal to the plane as Y. I asked him to clarify what is Y but it does not seem to be done yet. It is confusing if different people considers different Y-s.:smile:
 
  • #16
22,143
5,069
From the way he wrote the components initially (and the head tilting) it seems that the OP considers the direction normal to the plane as Y. I asked him to clarify what is Y but it does not seem to be done yet. It is confusing if different people considers different Y-s.:smile:

I agree. With every tilt of the head, he keeps flip-flopping. The OP seems to have no problem with his understanding of the physics. His issue seems to be with getting the geometry and trigonometry correct.
 

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