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Torque / Spring & Friction

  1. Jun 25, 2006 #1
    Alright guys..I got another couple of questions that I'm having trouble with..hopefully you guys can shed some light on em for me..

    1. A 85 kg scaffold is 6.6 m long. It is hanging with two wires, one from each end. A 500 kg box sits 2 m from the left end. What is the tension in the left wire?(g = 9.8 m/s2)

    I set the the right wire as the axis of rotation..
    Torque = 0
    (833x3.3N) + (4900x4.6) - T1 (6.6) = 0
    T = 7543.8

    EDIT: I just re-did this and got 3831.6..im assuming i originally just did the math wrong

    ---------------------------------

    6. A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.35, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?

    I have four forces acting on the ladder, FN1 & FN2, Fs, and Mg..

    [​IMG]

    First I broke down into components..
    X: fs = FN1
    y: FN2 = Mg

    Then I used Torque:
    Mg(L/2)sin(t) + UsFN2(L)sin(t) - FN2 (L) cos (t)

    I then essentially plugged in and solved for theta..and got 49.63

    ---------------------------------

    3. A mass m1 = 13.5 kg and a mass m2 = 11.5 kg are suspended by a pulley that has a radius of 10.4 cm and a mass of 2.8 kg (see figure). The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. Treating the pulley as a uniform disk, determine the acceleration of the two masses.

    For this I used a formula I remember the professor talking about:

    (Driving Force - Resting Force) / (Total Masses) + 1/2 (Mass of Pulley) = A

    I got .78..unfortunatley thats not right..: \

    [​IMG]

    ---------------------------------

    10. A 1.15 kg box rests on a plank that is inclined at an angle of 59° above the horizontal. The upper end of the box is attached to a spring with a force constant of 24 N/m, as shown in the figure. If the coefficient of static friction between the box and the plank is 0.24, what is the maximum amount the spring can be stretched and the box remain at rest?

    As for this one I don't even know where to begin..I'm assuming you can use energy conservation..but I'm not really sure what to plug in for x in the elastic PE or how to get the H in the regular PE..

    Any help appreciated..thanks guys

    [​IMG]
     
    Last edited: Jun 25, 2006
  2. jcsd
  3. Jun 25, 2006 #2
    Actually I just re-did my math on the first one and got 3831.6..im assuming i just did the math wrong
     
  4. Jun 26, 2006 #3
    1. I got the same answer as you.
    3. By any chance, is the answer 1.429m/s^2?
    Method:
    Moment of Inertia, I of an uniform disc = mr^2/2
    Torque = I * angular acceleration
    = I *( linear acceleration/radius)-----(1)
    Torque = 2N * radius ----- (2)
    Consider (1) and (2), i got the answer of 1.429
     
  5. Jun 26, 2006 #4
    To be honest..I'm not really sure what the answer is..for the third question..don't you have to consider the torque of both masses as well their forces?

    I re-did it and this is what I got..
    Torque = T1R - T2R = 1/2MR^2 * Ang Acc.
    F1 = M1G - T1 = M1A
    F2 = M2G - T2 = M-M2A

    Pluggin in I got:
    R(M1G - M1R * ANG ACC) - R (M2G + M2R * ANG ACC) = 1/2MR^2 * Ang Acc.

    I then got 7.15 for Ang Acc..

    And then used Ang Acc = a / r to find the a..for which I got .74..

    I'm still not sure if its right tho
     
  6. Jun 26, 2006 #5
    Ok so I think I got number three as well..I don't understand four at all tho..
     
  7. Jun 26, 2006 #6
    4) just use net force=0
    if spring is stretched "downward"
    mgsinθ-kx+friction=0

    if spring is compressed:
    mgsinθ+kx-friction=0

    edit: fixed some errors about the equation, sorry
     
    Last edited: Jun 26, 2006
  8. Jun 26, 2006 #7
    3) these types of question are best done considering the WHOLE system. it can be done in one line. since the whole system acceleration at the same linear acceleration.
    τ=torque
    α=angular acceleration

    the net force can be thought of the sum of all net forces of all bodies in the system.
    Fnet=Fnet1+Fnet2+Fnet3

    Fnet=m1a-m2a+τ/r
    Fnet=m1a-m2a+Iα/r
    α=a/r
    [itex]F_{net}=m_1a-m_2a+{I\over{r^2}}a[/itex]
    Fnet is the difference of the weight
     
    Last edited: Jun 26, 2006
  9. Jun 26, 2006 #8
    For the fourth one..where exactly did you get that formula from?

    So basically according to what you posted..

    9.8 * sin 59 + 24x - .24 = 0

    And then solve for x? What about the mass? Does the mass affect the situation?
     
  10. Jun 26, 2006 #9
    oop, sry, i forgot... yeah your right, its actually:
    if spring is stretched "downward"
    mgsinθ-kx+friction=0

    if spring is compressed:
    mgsinθ+kx-friction=0
     
  11. Jun 26, 2006 #10

    nrqed

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    Gold Member

    You forgot a mass in the first term. Also, the static friction force is NOT equal to .24! ).24 is only the *coefficient of static friction. The static friction *force* is [itex] \mu_s N [/itex] where [itex] \mu_s = 0.24 [/itex] and you will have to find the normal force, which is not mg (you are probably able to do that part, right? If not, ask).

    So the equation will be
    [tex] m g sin (59) +24 x - 0.24 N = 0 [/tex]

    patrick
     
  12. Jun 26, 2006 #11
    Ahh..i see..FN = mgcos(theta)..

    I got .34..sound about right?
     
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