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Homework Help: Normal Force Addition

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data

    How do you add the normal forces together after you derive the x and y component?

    2. Relevant equations

    Normal force = m*g*sinθ (y component)

    Normal force = m*g *cosθ (x- component)

    3. The attempt at a solution

    Bottom line is I'm trying to get the Net Force.

    Net force= mass of the object + normal force - kinetic friction force.
  2. jcsd
  3. Feb 17, 2012 #2


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    Welcome to PF!

    Hi Asuncion! Welcome to PF! :wink:

    (btw, mass isn't a force, you mean weight :wink:)
    (Are you talking about a body moving along a slope?)

    You don't "add the normal forces together",

    you add the component of one force to the same components of the other forces (in the same direction)
    No, those are the components of the weight, not of the normal force. :smile:
  4. Feb 20, 2012 #3
    I thought
    W (weight)= mg

    N (normal force)=Wcosθ

    Fk(kinetic friction) = μkN

    Here's the data given:

    mass= 10kg textbook
    μk = 0.3
    θ= 30
    g = 9.8 m/s2

    Find net forces. So I add the Weight, and Normal Forces together then subtract the kinetic friction force even though it lies on the positive x axis. Hummm......
  5. Feb 20, 2012 #4


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    Yes, that's correct. :smile:

    (the last line is a vector equation)
    That wasn't correct.
    That, I didn't really understand. :confused:
    I'm not following you.

    Add the x-components of all the forces, that gives you the net force (you know its y-component is zero, don't you?) …

    what do you get? :smile:
  6. Feb 20, 2012 #5
    Hey Tim! Thanks for your help. So here's what I've come up with after 3 hours of surfing different physics sites.

    Incline plane problems are more complex than other force problems because the forces acting on the object on the incline plane are not perpendicular. Therefore, components must be used (quote from Mr. fizix site.)

    Picture a box on a incline plane. W is the resultant vector. Therefore, you add the Wx and Wy components together.
    W= Wx + Wy
    Wx = Wsinθ
    Wy = Wcosθ

    I knew I must add the components together....I should have applied the basic vector addition. I was getting hung up over the "name".....So regardless if it's friction, weight, normal etc. As long as you have θ you can find the X and Y components.....I'm learn'in LOL....anyways so for the 100 kg small crate I got a net force of 1571. (pretty high though..hummm) and for the 10 kg textbook I got 154N, both having a acceleration of 15m/s2. Am I right? :/
  7. Feb 20, 2012 #6
    Sorry if I confused you I got the Normal force and Weight Force equations confused. Original Quote "How do you add the normal forces together after you derive the x and y component?" Now I know, add the Weight components not the Normal Force components.
  8. Feb 20, 2012 #7


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    sorry, you've lost me :confused:

    how did you get those figures?​
  9. Feb 20, 2012 #8
    Sorry. I promise I'm not trying to confuse you on purpose. I had to re-calculate. Here's what I came up with.

    m=100 kg
    Fn= 100 *9.8cos30=845 N
    Fw= 1339N
    W= 100 *9.8=980N
    Wx= 980cos30
    Wy = 980sin30
    (Add components together Fw=1339 N.)
    Fk =.3(845N)= 254N

    Fnet = Fn+ Fw - Fk=1930N
    a=19.3 m/s^2.
  10. Feb 20, 2012 #9
    Wait! Shouldn't the g be -g for the Fn equation because normal force is pointing upward against gravity? hummm.... hope I didn't confuse u even more :/
  11. Feb 20, 2012 #10


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    (try using the X2 button just above the Reply box :wink:)
    ah I see!

    no that's rubbish :redface:

    you can't add Wx to Wy

    you can't add any x-component to any y-component

    you can add all the x-components together

    you can add all the y-components together

    that's how vectors work …

    when you add them, you add the components (separately), not the magnitudes
  12. Feb 21, 2012 #11
    Okay Tim, help me out here. Honestly I have already turned in this lab, BUT I need to know the right way to solve the problem. It's almost like I'm about to breakthrough physics! I'm excited. So can ya help me out alittle here?

    Tim, How would you find the all the kinetic friction force, nornal, force, weight force, all finally the net force? and the acceleration? I ask because This information will be needed to solve the more complicated---deriving a formula that shows a relationship between the mass of an object and how far it travels off the end of the ramp (this is the ultimate tricky part.)

    And as I become more familiar with physics this is what's it all about. Adjusting the 3 main formulas to derive a New formula! Am I right! Kinda like solving in "more ways than one" anyways, how would you create a formula that shows a relationship between the mass of an object and how far it travels off the end of the ramp? Thanks Tim!!
  13. Feb 21, 2012 #12
    Oh yeah one more thing to address the "no adding x and y component vectors"......

    I wish I could attach the diagram I drew. :uhh: don't know how to do that, but any hoos, I was treating the weight force vector like the resultant vector. Therefore, according to pythagorean theorem you can add them together...hold that thought.......the normal vector and the friction vector are at a 90° angle does that matter!! Okay I'm officially lost.
  14. Feb 21, 2012 #13
    I'm going to start at the beginning to keep everything clear:

    An object of mass m sitting on a plane inclined at angle [itex]\theta[/itex]. In this situation we have to consider both the force of gravity and the normal force; however, the mass is sliding down the plane (if the mass is sliding up the plane a similar method follows) and so we have to consider the kinetic force of friction. So we have three different forces to consider in this problem.

    When solving inclined plane problems I find it easiest to think of vector components in terms of components parallel and perpendicular to the plane (you can use regular x- and y- if you're comfortable; you can actually use any co-ordinate system in which the axis are perpendicular but the different values for angles might be confusing; either way, I will stick to components parallel and perpendicular to the plane).

    I know this may be difficult to visualize, so if you're having trouble try drawing a diagram (an FBD) to help.

    The force of gravity points directly downwards. This force can be "broken up" into components such that you will have a parallel and perpendicular component of gravity. The perpendicualr component of gravity is the component Ipushing the mass into the plane; the parallel component is the component pushing the mass down the plane (this is what cause the mass to slide down the plane). So we can write the following:

    [itex]F_{g||} = mgsin\theta [/itex]

    [itex]F_{g\perp} = mgcos\theta[/itex]

    Convince yourself of this by drawing an FBD and considering only the force of gravity. Use similar triangles to show that [itex]\theta[/itex] is the angle of inclination.

    Note that these components will satisfy the Pythagorean Theorem such that:

    [itex]F_{g} ^{2} = F_{g||} ^{2} + F_{g\perp} ^{2}[/itex]

    Next consider the normal force. The normal force must act entirely perpendicular to the plane (there is no component of the normal force parallel to the plane); this may not always be true, but in this question it is (and pretty much every question I've dealt with. Explain why the normal force must act perpendicular to the plane. (Hint: Consider what would happen physically in the direction perpendicualr to the plane if the normal force was not equal to the perpendicular component of the gravitational force.)

    You can write:

    N = N[itex]_{\perp} = mgcos\theta[/itex]

    Lastly, consider the force of kinetic friction. The force of kinetic friction always acts opposite to the direction of motion; if the mass is sliding down the plane the force of kinetic friction points up the plane. What is the expression for the force of kinetic riction?

    Now we have all the information we need. You have found expressions for 3 forces and have written one of those forces in terms of its compoents so we have 4 expressions. If we perform a net force balance in both the direction perpendicular and parallel to the plane equating the sum of the forces to ma you should be able to solve. Perform a force balance in both the perpendicualr and parallel directions.

    Does this make sense? Please ask if you have any questions (if anyone notices a mistake, please correct me; I was a bit sloppy in terms of positive and negative directions parallel to the plane).

    Also, it helps to solve everything in terms of variables instead of pluggin numbers so you can see what your final equation looks like. Once you have an equation you can check its units to see if what you've done is (probably) correct.
  15. Feb 21, 2012 #14


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    Hi Asuncion! :smile:
    you start with whatever the question gives you, and then apply the formulas

    if a formula says eg F = ma, sometimes you know F and m and you have to find a, sometimes you know a and m and you have to find F, sometimes you know F and a and you have to find m :wink:

    if there's more than one formula to use, you just have to choose the correct order to link them all together so that you keep using what you already know, and finding something to use in the next formula
    that would be conservation of energy … better wait until you do that in class! :wink:
    it depends what you mean by "add"

    scalars (ie, ordinary numbers) add like scalars

    vectors add like vectors (as in a vector triangle or vector parallelogram)

    for three vectors A B and C, you can have a vector addition equation A + B = C

    or three scalar addition equations

    Ax + Bx = Cx

    Ay + By = Cy

    Az + Bz = Cz

    and you can even add Axi Ayj and Azk to give the resultant A (which you're calling pythagorean addition) :smile:
  16. Feb 21, 2012 #15
    "Conservation of Energy"--yeah might wanna wait on this tuff cookie.

    Okay guys let's get this right. There's an object (small crate) that's on an incline plane that's at a 30° angle. The object (small crate) weighs 100 kg. The incline is made of wood so the kinetic friction is 0.3. Okay. With that given information I am going to find the normal force. The kinetic friction force and the gravity (a.k.a weight) force. Okay lets began.

    Normal force Fn=m*gcosθ
    Fn=100 kg*9.8 m/s^2cos30=848N

    Friction Force Fk=μn

    Here we go again..

    Weight force=Wsinθ

    Now let's add'em Up..
    Fnet = Fn+ Fw - Fk=848N + 490N - 255N= 1083N

    Please tell me I did this right...If not just show me how please.
    Last edited: Feb 21, 2012
  17. Feb 21, 2012 #16
    Oops I left out how I got the weight.
    W=100 kg * 9.8 m/s^2=980N

    Oh and I only used the Wy component of weight. Not the Wx component of weight. Thanks. Again if I didn't do this right just show me how. Thanks guys.
  18. Feb 21, 2012 #17
    Can I use microsft to draw FBD? or is there another program/software you can recommend? thanks ;)
  19. Feb 21, 2012 #18


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    Hi Asuncion! :smile:
    No!! you've done it again! :redface:

    you've added perpendicular forces.

    That equation is for the x-components only, there's no normal force there.

    Useful tip:

    Stop using "F" for force …

    haven't you noticed, they're all forces! :rolleyes:

    Use "N" for normal force,"W" for the weight, "F" for the friction force,etc

    And then you can write eg Wx + Fx = 0 …

    that way you won't get y components in an x equation!! :wink:

    Try again. :smile:
  20. Feb 22, 2012 #19
    So are you saying that the net force only consist of the forces that's parallel to the x axis. If so than the normal force and weight force cancel out and that just leaves the kinetic force. Am I right? If so, the net force=-255N. it's negative because its going in the opposite direction.
    Last edited: Feb 22, 2012
  21. Feb 22, 2012 #20


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    Have you drawn a diagram? :confused:

    How can the normal force have anything to do with the x direction?
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