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Mass on an Incline

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A block with mass m1 = 9.2 kg is on an incline with an angle θ = 25.0° with respect to the horizontal. For the first question there is no friction between the incline and the block.

    The coefficient of static friction was deterimined to be μ = .121

    The premise of this question is that there was a spring holding the box up(the spring only exerting force in the X direction), but this spring was replaced by a STRING that is parallel to the horizontal, thus making a 25° angle with the incline.

    Word for word, it is:
    The spring is replaced with a massless rope that pulls horizontally to prevent the block from moving. What is the tension in the rope?

    2. Relevant equations
    Fgx = mgsin(theta)
    Fgy = mgcos(theta)

    Fty = Ftsin(theta)
    Ftx = Ftcos(theta)

    Ff = μN

    N = mg + Fty

    3. The attempt at a solution
    This problem has been frustrating me a lot, I've probably been stuck for hours on this one part on my already done homework... I just want to know how it's done because I'm definitely not going to be able to get credit at this point...

    I've figured that Ff = 9.886 + μ(Fty) and so the force of friction is dependent on the tension force in the Y direction. Also, the tension force is dependent on the force of friction because Ftx = Fgx - Ff. It seems impossible for me because every time I try to setup a systems of equations, I end up with 0 = 0, which doesn't really help me. I've tried dozens of approaches, this one seems to be the closest because it factors in the tension in the Y direction into the normal force, which incorporates into the frictional force, but I can't figure it out.

    A solution would be greatly appreciated.
     
    Last edited: Sep 25, 2012
  2. jcsd
  3. Sep 25, 2012 #2
    Have you tried to draw free body diagrams of the block in both scenarios? A general diagram of the whole setup will assist you for this purpose (and in general).
     
  4. Sep 25, 2012 #3
    Yep, and determined that the only two forces acting UP the incline is friction and tension force(the parallel X component), and also that there is a Y component of tension perpendicular to the slope that would increase the normal force, which would increase the friction, which would reduce the tension. It's just a confusing loop and I can't seem to setup the system of equations(which I presume is the only way this can be solved)

    BTW I corrected myself, it's coefficient of static friction, not kinetic.
     
  5. Sep 25, 2012 #4
    I am not sure if I am picturing this correctly, so I drew how I understand the situation. Correct me if that is wrong.

    If not, then there is no component of tension force perpendicular to the incline, at the box. Notice that I have drawn all the forces on the box.
     
  6. Sep 25, 2012 #5
    Nope, the string is connected in such a way that it is parallel to the HORIZONTAL, so it makes an angle of 25 degrees with the incline.

    So there is a component parallel to the incline(Tcos(25)) and a component perpendicular to the incline that has to be factored into Friction/Normal force calculation (Tsin(25))
     
  7. Sep 25, 2012 #6
    A clarification: In that picture, the dotted line is the string, so it is parallel to the horizontal.

    Can you draw the correct setup?
     
  8. Sep 25, 2012 #7
    Here's the correct setup
     

    Attached Files:

  9. Sep 25, 2012 #8
    Ah, I see. It seems you have solved the problem in your first post. Below I have quoted it indicating the substitutions you need to make (replace darker shade by lighter shade of same color). I do not get a 0 = 0 so I'm guessing you made some algebraic error. Do it again and see if you still get such discrepancy and post it here if you do.

     
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