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Mechanics Problems: Tensile Forces and Pressure

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A uniform, horizontal beam of length 4 weighs 150 N. It is connected to a wall by a hinge and is held horizontal by a cable of length 5; it holds a 300 N weight at its end away from the wall. The cable and the beam form a 3:4:5 triangle with the wall.

    2. Relevant equations

    [tex]\sum_{i=1}^n\textbf{F}_i=m\textbf{a}_i=0[/tex], because the system is not accelerating.

    3. The attempt at a solution

    The forces acting upon the end of the rod are half of its weight, the hanging weight, and the tension. I've solved for [tex]textbf{T}[\tex] to no avail. What am I doing incorrectly?
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2


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    Hi asleight! :smile:

    Show us what you tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Nov 9, 2008 #3
    Examining the FBD for the end of the rod, we see:

    [tex]\sum\textbf{F}=\textbf{F}_{g,weight}-\textbf{T}\sin{\theta}=0[/tex], solving for tension, we find:

    [tex]\textbf{F}_{g,weight}/\sin{\theta}=\textbf{T}[/tex], is that really it?
  5. Nov 9, 2008 #4


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    Nooo … F and T are vectors, and they're not parallel, so that can't be right can it?

    You've left out the unknown force at the hinge.

    Since it's unknown, what technique do you know that will give you an equation without it? :smile:
  6. Nov 9, 2008 #5
    I don't see what the problem is. Parts (b) and (c) ask to solve for the forces applied by the hinge so I don't know where to go on from here. I'm not so intuitively-savvy today, it seems. I have other questions, too. GAH. Haha.
  7. Nov 9, 2008 #6


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    Try taking moments about the hinge.
  8. Nov 9, 2008 #7
    Solving for the sum of the torques, we find 1200Nm being applied in the clockwise direction by the hanging mass, 300Nm by the rod itself, also applied in the clockwise direction. Then, we have to cancel that with the torque due to the chord. So, the torque is 1500Nm.

    [tex]\tau=r\times F=rF\sin{\theta}=3rF/5=1500[/tex],

    solving for F, we find: F = 625 N, is that right?
  9. Nov 9, 2008 #8
    Nevermind, I finished that problem. The next one'll be posted shortly.
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