Inclined planes + vector questions

  • Thread starter Rhine720
  • Start date
  • #1
88
0
So, my physics teacher has made this VERY confusing.
We use vectors. But for some reason I keep wanting to think the resultant vector is the ultimate vector for magnitude and direction of force,speed acceleration w/e. But for some reason I'm not seeing that

Normal force is the force that pushes up,right? Like if I'm standing on the ground, the ground is exerting normal force on me? (For equil with gravity)

Anyway with my inclined plane, it says to drop a gravity vector, down and then to work my trig skills to finisht he triangle. so why would that hypotenuse be useful since the direction would be the line parallel to the sloped part of the inclined plane?

Also When we break vectors into components. We're saying, its moving along this axis this much and that one that much? Like, A slanted line is moving long the x at 4 but also along the y at 3, causing a slope..right(the resultant?)
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,165
507
So, my physics teacher has made this VERY confusing.
We use vectors. But for some reason I keep wanting to think the resultant vector is the ultimate vector for magnitude and direction of force,speed acceleration w/e. But for some reason I'm not seeing that
well, you are right, the magnitude of the resultant vector is greater than (or equal to ) any of its component parts.
Normal force is the force that pushes up,right?
it pushes perpendicular to the surface on which it acts, but not necessarily 'up'
Like if I'm standing on the ground, the ground is exerting normal force on me? (For equil with gravity)'
yes, this is correct
Anyway with my inclined plane, it says to drop a gravity vector, down and then to work my trig skills to finisht he triangle. so why would that hypotenuse be useful since the direction would be the line parallel to the sloped part of the inclined plane?
the hypotenuse becomes the weight force, mg, acting down; you can break the resultant weight force into its components acting along the plane and perpendicular to the plane, using geometry and trig. This is very useful in determining the normal force and acceleration along the plane.
Also When we break vectors into components. We're saying, its moving along this axis this much and that one that much? Like, A slanted line is moving long the x at 4 but also along the y at 3, causing a slope..right(the resultant?)
It may not be moving, though....but you have the right idea....now when you do inclined plane problems, it is very convenient to choose the x axis parllel to the incline, and the y axis as perpendicular to the inclne...in this manner, you can oft readily solve for the normal force, because you generally can use newton 1 in the direction perpendicular to the incline, but you can't use newton 1 in the vertical direction.
 
  • #3
88
0
Thanks. Well I'm looking at my own drew up inclined plane thingy with a 200kg box on it. I know that if i draw a light straight down Il'll get the weight and hypot of another triangle. So what good is that hypot? isn't it a resultant vector already?
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,165
507
Yes, but its component perpendicular to the incline can be equated to the normal force (N= mgcostheta); and its component parallel to the incline (mgsintheta) gives the component of the weight which acts down the incline, which yields an acceleration down the incline in the absence of friction and other forces.
 
  • #5
88
0
hmm..but wouldn't getting the normal force first work? Thengetting weight and using pythag rearranged to get the last bit work?
 
  • #6
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,165
507
hmm..but wouldn't getting the normal force first work? Thengetting weight and using pythag rearranged to get the last bit work?
Please show the equation you would use to get the 'normal force first?'.
 
  • #7
88
0
If i drew a vector perpendicular to the object? Then 9.8m/s^2*200kg
 
  • #8
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,165
507
If i drew a vector perpendicular to the object? Then 9.8m/s^2*200kg
Well, no, if you drew a vector perpendicular to the object, that could be a component of the weight, not the weight itself. The weight is mg acting vertically down; its component perpendicular to the plane would be less, and its componet parallel to the plane would be less. The vector sum of those 2 components is the weight. There is another force acting perpendicular to the object besides the component of the weight; it is the Normal force. Once you find the weight component perpendicular to the plane, can you find the normal force, using Newton's laws?
 
  • #9
88
0
Oh. That makes so much more sense, because gravity only works straight down (to the middle of earth).So When i draw a vector going straight down for the hypotenuse I get that the object is being pulled down at this rate, and because of newtons laws or normal force, its not falling straight down, so its sliding downward? Like sau, 980 n pushing down the slope of the plane. y amount of newtowns pushing it down and x amount of newtons pushing it to the left. Because of normal force though it' doesn't fall through the plane. soo uhm?

Can i see like a diagram? I've tried making my own to see if i have the concept, but i can't quite do that.
 

Related Threads on Inclined planes + vector questions

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
972
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
2K
Top