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2 more basic problems

  1. Feb 16, 2008 #1
    [SOLVED] 2 more basic problems

    3 dogs are pulling a sled with a combined weight of 145 kg the dogs are pulling
    1) pulling at a force of 83N at 15.5(degrees) left of forward
    2) pulling at a force of 75N at 9 (degrees) right of forward
    3) pulling at a force of 77N at 12(degrees) right of forward

    sled is moving a a constant velocity find the coefficient of kinetic friction between the sled and snow,
    so the sled has a force of 1422.45N in the opposite direction the dogs are pulling, so then i have to find the vertical forces the dogs exert and add them together? so to find vertical i use f.sin(theta) then add the 3 together so i know.
    1) Fy=22.18N
    2) Fy=11.73N
    3) Fy=16N

    i dont even know if im going about this the right way,

    ill give problem 2 once i solve this
  2. jcsd
  3. Feb 16, 2008 #2
    actually on thinking of it i was using Fg to find the Ff but now im stuck as to how im finding the Ff
  4. Feb 16, 2008 #3
    You are off to a good start, but be careful of which trig function you use. Now what does constant velocity imply in terms of acceleration and forces?

    Ff = -µN

    where N stand for the normal force.
  5. Feb 16, 2008 #4
    constant velocity implies no acceleration which means net force is 0 but to find µ i need Ff do I not? did i do the forces of the dog right?

    Fn would equal 1422.45 correct
  6. Feb 16, 2008 #5
    The forces from the dog are wrong because of a subtlety you missed, think about where the angles are defined from. Right, constant velocity implies no acceleration, so the forces from those dogs better cancel the force from friction.
  7. Feb 16, 2008 #6
    so we can say Ff = whatever the F of the dogs combined is or the - of it, and i dont know what i missed on the forces is it supposed to be sin? or the angle part should i be starting my measurements at E ? anglewise

    this gives me
    1) force of 79.98
    2) force of 74
    3) force of 75
    Last edited: Feb 16, 2008
  8. Feb 16, 2008 #7
    Oh, well maybe you are right, but your numbers are wrong. Fdog*cosØ = Fforward

    Yes, Ff = Fdogs because we know that Fdogs+(-Ff)=0.
  9. Feb 16, 2008 #8
    so force dog cosØ is the same as the sin i mentioned when i edited my post when measuring off east? nice so FF=229 so i can go mu k equals 229/1422.45 which gives me .16 the book is saying .15x10(2) im confused
  10. Feb 16, 2008 #9
    The dogs should have a total force of 80+74+75=150+80-1=229, might as well say 230 for niceness. You are right .16. What does your book say? 15? What??
  11. Feb 16, 2008 #10
    yea the book says 15 but whatever if you agree with me thats all that matters.
  12. Feb 16, 2008 #11
    Well coefficients of friction don't exceed 1, so your book is screwy.
  13. Feb 16, 2008 #12
    alright and the other problem i had was simple it is a pulley question where 2 blocks are connected block one is on the incline and weighs 47kg and block 2 is on the other end of the pulley and weighs 35kg, theres 4 parts but the only one i need help with is a will the blocks move if m2 is released, i know the answer is NO but how would you figure that out formula wise i should know this but worth asking

    heres a diagram


    red is Mass 1 Black is Mass2 and they move on a pulley, quite crappy 2 minute diagram
  14. Feb 17, 2008 #13
    i know you have to see if fa is greater then Ff but i dont know what formulas to use..any help
  15. Feb 17, 2008 #14
    anyone, id love to get this unit done with today, last question to do, still stumped and looked back through the textbook
  16. Feb 17, 2008 #15
    Draw a free body diagram for block A. There is a force of m2*g pulling to the right, and to the left a force of m1*g*sin(25) from the weight of the block, plus whatever friction provides, ie mu*m1*g*cos(theta).
  17. Feb 17, 2008 #16
    so i do all that and if they equal the same the system wont move correct? so i know Fn is 417.87 for block 1 and i know the right force is 343.35N so to find left components i can go Fleft + Ff so 194.85 +175.5 that equals 369 though which would mean it would move down the ramp
  18. Feb 17, 2008 #17
    what is the coefficient of friction?
  19. Feb 17, 2008 #18
    muk is 0.19 and mus is 0.42
  20. Feb 17, 2008 #19
    for block 1:
    m1*g*sin(theta)+mu*m1*g*cos(theta) - T = m1*a

    For block 2:
    T - m2*g = m2*a

    eliminating T and solving for a
    a=g*(m1*sin(theta) + mu*m1*cos(theta) - m2)/(m1 + m2)

    The only way the system stays at rest is when
    m1*sin(theta) + mu*m1*cos(theta) - m2 = 0
  21. Feb 17, 2008 #20
    so my book is saying that it stays at rest? is it correct,

    because thats not the answer i am getting
    Last edited: Feb 17, 2008
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