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Homework Help: Running out of Time, Please help me !

  1. Dec 5, 2004 #1
    ok, I have my assignment due by noon today and it is now 2:00am (I've been up most of tonight trying to get 4 questions done and they are all driving me insane) Of the 4, 3 are dealing with static Equilibrium So if you can Help with ANY of them I'll really appriciate it.

    1) A rod of length 68.0 cm and mass 2.00 kg is suspended by two strings which are 38.0 cm long, one at each end of the rod. The string on side B is cut. Find the magnitude of the initial acceleration of end B.
    A diagram can be found here: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob13a_1016full.gif

    my workings:

    net force = 0
    A + B - mg = 0

    net torque = 0
    picking pivot to be at point A
    LB - mgR = 0

    so solving the torque equation to find
    B = (9.8)(2)(0.34) / (0.68)
    B = 9.8N

    so A = 9.8N as well (solving the net force eqn)
    I'm not even sure if I have that much done right, or even if i had to do that, I guess the length of the strings comes into play somewhere but I haven't used them yet, I need a lot of help with this one I think.

    2) A 28.0 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of theta=13.1° with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam?
    a graphic is here: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype24/prob19a_beamhinge1.gif

    so my workings

    Ty + Fy - mgsin(90-13.1) = 0
    Fx - Tx = 0

    LTy - L/2 mgsin(90-13.1) = 0
    Ty = (9.8)(28)(sin(76.9) / 2
    Ty = 133.63 N

    Tsin76.9 = 133.63N
    T = 137.20

    Tcos76.9 = Tx
    Tx = 31.1 N

    Fx - 31.1 N =0
    Fx = 31.1 N

    And this is wrong I know, I really dunno where I've gone wrong but I only have 3 more tries left on this question so I'm getting a little stressed out by it.

    3) A sledgehammer with a mass of 2.40 kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_cm=0.570 m, and the moment of inertia about the center of mass is I_cm = 0.0360 kgm2. If the hammer is released from rest at an angle of theta=59.0 degrees such that H=0.489 m, what is the speed of the center of mass when it passes through horizontal?
    A graphic: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob34_sledgehammer.gif

    I'm really not sure what to do with this one.... I'm getting really tired and stressed and my concentration is slowly dying...

    4) A solid door of mass 33.10 kg is 2.40 m high, 1.00 m wide, and 2.94 cm thick. If the edge of the door has a tangential speed of 45.7 cm/s, what is the rotational kinetic energy of the door?

    I've already found the Moment of inertia for the door about an axis through it's hinges to be 11 kg*m^2 but now I'm not sure where to go with it?

    I know I haven't explained my own workings very well, but like I said it's 2am here and I can't really concentrate very well. I really need help, I've been trying to figure out these last 4 questions for a couple days now, the rest of the assignment was easy compared to these.
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Dec 5, 2004 #2


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    In (1), you have determined the forces (A and B) on the rod before the string is cut. Now when B is cut, the force at B becomes zero. At that instant, the force on A and the weight are the only forces acting on the rod.
    From these unbalanced forces, you can find the acceleration of the CoM and from the unbalanced torque, you can find the angular acceleration about the CoM. From these 2 numbers you can get the initial accelration at the end B.
  4. Dec 6, 2004 #3


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    (2) : Instead of resolving forces horizontally and vertically, it's a lot easier if you resolve them parallel and perpendicular to the rod. This way you only have one force to resolve. Don't panic...that'll only make things worse.

    This is really a simple problem, compared to the first one, unless I'm missing something obvious (which I suspect I am) on that one.
  5. Dec 6, 2004 #4


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    (3) Use energy conservation. Initially you have some potential energy with respect to the level of the pivot. Finally, you have rotational KE about the pivot, and no PE. Equate the two to find the final angular velocity, and from that, the speed.
  6. Dec 6, 2004 #5


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    (4) You've found the moment of inertia about the hinges. That's the first step. Now what's the formula for the rotational KE ? Besides the moment of inertia, I, what other quantity do you need for calculating this ? How do you determine this quantity from the other piece of data ?

    PS : Judging from 2,3, and 4, it looks like I may be suggesting a roundabout approach for solving 1. I would suggest you try 2, 3 and 4 first.
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