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Physics hw. our teacher offered us 5 extra pts on our test on mon. if we get c right

  1. Nov 10, 2009 #1
    1.AB is a uniform bar of length 5m and weight 60N. It is supprted in a horizontal position by 2 vertical strings P and Q. P is 0.5m from A
    and Q 1.5m from B.

    a) What are the tensions in the strings?
    b) What weight must be applied at A if the tensions in the strings are to be equal?
    c) What is the biggest mass which can be hung from B if the bar is to remain horizontal?


    please help....i got a) but im not sure about b and c
     
  2. jcsd
  3. Nov 10, 2009 #2
    this h.w is for 5 extra points on a test on monday if i get them right...please help.

    1.AB is a uniform bar of length 5m and weight 60N. It is supprted in a horizontal position by 2 vertical strings P and Q. P is 0.5m from A
    and Q 1.5m from B.

    a) What are the tensions in the strings?
    b) What weight must be applied at A if the tensions in the strings are to be equal?
    c) What is the biggest mass which can be hung from B if the bar is to remain horizontal?


    a) M(P) :

    3*q -2*60 = 0
    3q-120=0
    q=40

    [resolving vertically upwards]:
    p+q-60=0
    p+40-6=0
    p-20=0
    p=20
    ********************************************************************
    b) finding P:

    60-4.5p+5F=0
    (-40 + 2p-2.5F=0)*2

    60-4.5p + 5F= 0
    -80 + 4p-5F=0

    -20 + .5p = 0
    .5p = 20

    p = 20/.5
    p= 40N


    finding F:

    40-60+40-F = 0
    20-F=0
    F=20N
    ******************************************************************

    M(F):
    B-40+60-40+20=0
    B= 0

    ????? i don't know if these are right...i tried. Please help me.
     
    Last edited: Nov 10, 2009
  4. Nov 10, 2009 #3
    Re: physics hw. our teacher offered us 5 extra pts on our test on mon. if we get c ri

    You draw all the forces and in each case you write down the following two equations:
    Total force up=Weight
    Total torque counterclockwise about center of mass=0
    (you have to pay attention to the *signs* since the forces are taken "up" as positive and the torque "counterclockwise" about the center as positive [you can also chose clockwise if you like, but at least all torques in the *same* direction about the center of mass])
    The torque is (distance from center of mass) times (perpendicular force in counterclockwise direction). For example if the force on string P is called P and string P is 2m from the center of mass, then the torque is -2*P (negative since its not counterclockwise as we agreed). Likewise Q is 1m from the center so the torque is 1*Q.

    a)
    forces and torque equations:
    P+Q=60
    -2*P+1*Q=0
    Solve this set of equations and get P=20N and Q=40N

    b)
    again forces and torque this time with a new force A at point A (and distance 2.5 from center):
    P+Q-A=60
    2.5*A-2*P+1*Q=0
    and P and Q should be equal so
    P=Q
    Solving these equations (use P=Q first to get rid of Q and continue) you get
    A=15N and P=Q=37.5N

    c)
    this time the extra force is B at B:
    P+Q-B=60
    -2*P+1*Q-2.5*B=0
    for the bar to stay horizontal you want string P to pull and not push (which a string cant do). so
    P>0
    canceling Q in the first two equations you get an equation for P and pluging this in into the last you get
    20-0.5*B>0
    This corresponds to
    B<40N
    which is the maximum weight you can apply at B.
     
  5. Nov 15, 2009 #4
    Re: physics hw. our teacher offered us 5 extra pts on our test on mon. if we get c ri

    Thanks so much :D:D:D
     
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