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    Echo901's Avatar
    Echo901 Posts: 6, Reputation: 1
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    #1

    Nov 6, 2013, 06:50 PM
    Physics: Torques
    A 1350-N uniform boom at phi = 64.5° to the horizontal is supported by a cable at an angle θ = 25.5° to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight w = 1850 N hangs from its top.
    Here is a link for the diagram: http://www.webassign.net/sercp9/8-p-028.gif

    Find the tension in the support cable (kN)
    Find the components of the reaction force exerted by the pivot on the boom. (kN)
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Nov 7, 2013, 06:31 AM
    Hint: start by using sum of moments about the pivot point to find tension in the cable. The rest is then using sum of forces in the x- and y-directions equal 0.
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    Echo901 Posts: 6, Reputation: 1
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    #3

    Nov 7, 2013, 05:34 PM
    Quote Originally Posted by ebaines View Post
    Hint: start by using sum of moments about the pivot point to find tension in the cable. The rest is then using sum of forces in the x- and y-directions equal 0.
    so I found the tension in the cable to be 1.45kN
    and using the following equation I found Fy to be 2.58kN
    Fy(Beam)+Fy(weight)-F(tension)*sin25.5

    But now I am a little confused as to how to set up the equation to get Fx. I know that Fx=0 and the components that make up Fx are the x components of the beam, the weight and the tension in the cable, but I don't really know how to approach setting up the equation, and I am unsure of which angle to use as well.
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    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Nov 8, 2013, 06:58 AM
    The x-component of the weight and the boom are both zero, because the force of gravity is straight down (in the y-direction only) So you have:

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    Echo901 Posts: 6, Reputation: 1
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    #5

    Nov 8, 2013, 07:52 AM
    Thank you so much!

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