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  1. J

    Pully and Inclined Ramp

    Oh, I assumed the two blocks would have the same acceleration as they are all in the same system... Shouldn't acceleration of A be equal to acceleration of B?
  2. J

    Pully and Inclined Ramp

    Nope.. it shouldn't be.. it should be equal to Ma A, not 0
  3. J

    Pully and Inclined Ramp

    I'm guessign at that point I could then use: Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ? Tension2*cos(22) - Magsin(30) - ( - Tension sin(22) + mgcos(30) )μ = 0 Put this equation in terms of Tension2 solve the hanging block equation for t1 Mb a = T1 - Mb g Tension1= Mb a + Mb g then...
  4. J

    Pully and Inclined Ramp

    Okay gotcha.. so I guess I can just account for the Ay as 0, meaning my normal force equation would be Normal Force = - Tension sin(22) + mgcos(30) as the 0 would cancel out the mass
  5. J

    Pully and Inclined Ramp

    oh.. I mean it would be 0, if the crate is remaining on the horizontal floor
  6. J

    Pully and Inclined Ramp

    I suppose I could just say Asin(22) = Ay... I'm not sure at this point.. this problem has been so drawn out to the point wher ethis isn't helping much anymore..
  7. J

    Pully and Inclined Ramp

    The y component would be sin() of it... the only thing i am confused about now Now I have ay and ax though? Look I really appreciate your help but going through this step by step has become pretty tedious, what do you think the equation should look like?
  8. J

    Two Blocks and a Pulley Friction Problem

    Homework Statement System comprised blocks, a light frictionless pulley and connecting ropes (see diagram). The 9.0kg block is on a perfectly smooth horizontal table. The surfaces of the 12kg block are rough, with μk = .2 between the two blocks. If the 5.0 kg block accelerates downward when it...
  9. J

    Pully and Inclined Ramp

    The block is going to accelerate up the ramp... I guess to calculate Ay at this point I would use something like this: Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?
  10. J

    Pully and Inclined Ramp

    This would be my equation for may, may * acceleration = Normal Force + Tension sin(22) - mgcos(30) I guess I could only solve for the normal force within the whole system of equations, or I could put this in terms of the normal force and solve for this as Normal force in the other equation? If...
  11. J

    Pully and Inclined Ramp

    Okay, so here is what I think the force diagram is for the block, ignore the other stuff though. So my equation for part 1 is Tension*cos(22) - Ma g sin(30) - ( Ma g cos(30) - Tension*sin(22) )μ = 0 sometimes I think about it such as the equation is parallel and perpendicular. Would this be...
  12. J

    Pully and Inclined Ramp

    Could I instead use my original equation I used to solve for mass b instead? very lost..
  13. J

    Pully and Inclined Ramp

    Otherwise.. gonna need some help here.. really not sure what the Y component could be
  14. J

    Pully and Inclined Ramp

    This was my Y component: 100gcos(30) - (Tension2*sin(22 What should the Y component be then? a little lost.. the axis are tilted.. should it be 100gsin(30) - Tension2*cos(22)
  15. J

    Pully and Inclined Ramp

    Oh, okay.. i see what your saying.. wouldnt it be something like: 100gcos(30) - (Tension2*sin(22) + 100gcos (30) The tension and normal force acting in one direction, with the force of direction in the other? other wise I am a bit lost here haha Thanks
  16. J

    Pully and Inclined Ramp

    Hmm, not quite sure what that means, would it be: (100gcos(30) - Tension2*sin(22))/Ma= A
  17. J

    Pully and Inclined Ramp

    Sorry, Not sure what I ment by that. I put 100gcos(30) - Tension2*sin(22) - Ma A in because I figured that when I calculated the normal force I would need to account of Acceleration as well since it is now moving.. now that I say that it doesn't make a ton of sense to me.. should I have just...
  18. J

    Pully and Inclined Ramp

    Woops, Thanks for the fast reply. Yeah, that's what I was implying. Looking at the diagram, I was trying to account for the Y side.. and figured that the Y would be changing as well as it is not at rest.. I edited the original post, I stupidly forgot the G. Tension2*cos(22) - Magsin(30) - (...
  19. J

    Pully and Inclined Ramp

    Homework Statement 100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The...
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