1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Blocks with a 'x' force and a 'y' acceleration

  1. Jun 22, 2013 #1
    1. Two forces are the only forces acting on a 2.8 kg object which moves with an acceleration of 3.8 m/s^2 in the positive y direction. One of the forces acts in the positive x direction and has a magnitude of 6.5 N.What is the magnitude of the other force f2?



    2. Relevant equations
    F=ma


    3. The attempt at a solution

    I suppose that to get the second force i just have to multiply F[SUB/]2[SUB/]=ma=(2.8kg)(3.8m/s^2) right?
     
  2. jcsd
  3. Jun 22, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    No. In the equation F = ma, F is the net force (the vector sum of all of the forces acting on the object). For a problem like this, it will be a good idea to express the second law in component form:

    ƩFx = max
    ƩFy = may
     
  4. Jun 22, 2013 #3

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    [Took the liberty of fixing up some format errors.]

    That gets you almost half way there. :smile: But not all of the way there. :frown:

    Is there a non-zero component of the object's acceleration in the x-direction? What does that say about the sum of the forces in the x-direction?

    [Edit: Tsny beat me again. Serves me right for typing too slow.]
     
  5. Jun 22, 2013 #4
    so basically, the whole thing is accelerating in a sum vector ? that i have to find with f2=sqrt(fx^2 + fy^2)?
     
  6. Jun 22, 2013 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. See if you can find f2x and f2y.
     
  7. Jun 22, 2013 #6
    well, I already have fx and now i have to get fy= m*ay and then i just substitute the values of fx and fy in f2=sqrt(fx^2 + fy^2) am I correct?
     
  8. Jun 22, 2013 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think so. Just to make sure you're going through the reasoning correctly, the equation ƩFy = may would be expressed as f1y + f2y = may, right? But then you already know the value of f1y, so you can find f2y. I believe that's how you are thinking, just wanted to make sure.
     
  9. Jun 22, 2013 #8
    no! i had another reasoning. i was going to just get fy= ma=(2.8 kg)(3.8m/s^2) and then because they give me already fx = 6.5 N ia was just going to get the vector sum, F= sqrt(fx^2 + fy^2)
    it was very different!
     
  10. Jun 22, 2013 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Oh, not good. There are two forces acting on the object: f1 and f2. Each force is a vector. You need to find the magnitude of one of those forces, namely |f2| = √(f2x2+ f2y2) .

    So you need to find f2x and f2y. You are not given either of those. But you can find them by setting up ƩFx = max and solving for f2x and setting up ƩFy = may and solving for f2y.
     
  11. Jun 22, 2013 #10

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    I think you are both talking about the same thing. TSny's notation is less ambiguous though.
     
  12. Jun 22, 2013 #11
    oh! yes, we're talking about the same thing, but the only thing that i didn't quiet understood in TSny's notations were the f2's. i thought that TSny's was trying to tell me that i had to get 2 forces which were f1 and f2 and each one was fx1=m*ay and f1=m*ax, and for fy2=m*ay and fx2= m*ax. but know i see that he's just telling me that:

    |f2| = √(f2x2+ f2y2)

    f2x= Fx /m = ax
    f2y= Fy=m*ay

    right?
     
  13. Jun 22, 2013 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I'm afraid that's not right at all. Think about it...How can a force equal an acceleration? They don't even have the same dimensions (units). Also, it's not clear what the symbols Fx and Fy mean here.

    Let's take it kind of slowly. Consider the fundamental equation ƩFx = max. Can you write out the left hand side of the equation using one or more of the symbols: f1x, f2x, f1y, and f2y?
     
  14. Jun 22, 2013 #13
    f2x-f1x=m*ax ?
     
  15. Jun 22, 2013 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    That's close, but why the minus sign on the left? The notation ƩFx means to add together all of the x-components of all of the forces acting on the object. That gives the total x-component of force acting on the object and it's that total x-component of force that equals max.

    So, what should the equation ƩFx = max look like?
     
  16. Jun 22, 2013 #15
    fx1+fx2=m*ax
     
  17. Jun 22, 2013 #16

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Very good. Now, what is the unkown quantity in this equation for which you want to solve? What are the numerical values of all of the other quantities?
     
  18. Jun 22, 2013 #17
    dude! i already solved this equation ! i just did the following thing:

    f2=sqrt((f given)^2 + (mass*acceleration y)^2)

    i got it right
     
  19. Jun 22, 2013 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Getting the right answer is one thing, clearly understanding the solution is another. Do you feel that it is all clear now?
     
  20. Jun 22, 2013 #19
    i think that what you were trying to teach me was deriving all the formulas to get what i did right?
     
  21. Jun 22, 2013 #20

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes, I just wanted to make sure you understood the thought process.

    For example, suppose f1 was still 6.5 N but in a direction of 30o above the positive x-direction. Everything else in the problem is kept the same. Would you still be able to solve for the magnitude of f2? If so, then I think you are understanding it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Blocks with a 'x' force and a 'y' acceleration
Loading...