Recent content by FairyChatMan

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    Friction of Two Blocks and Rope Problem

    Homework Statement Two blocks are positioned according to the figure. Block A (80 lb) is attached to a rope as such that it doesn't slide or fall from block B (120lb). Compute the smallest value of Force to be applied on Block B for it to not move parallel to the inclined plane...
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    Truss Analysis: Find Forces on Members

    Truss Analysis :( Homework Statement Find the force on each of the members and tell if it is tension of compression. Homework Equations M = fd \sumFx and Fy = 0 The Attempt at a Solution Well, my professor said to us that the first thing to do is to compute for the...
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    Solving Billboard Problem with Equations of Motion

    ah.. ok I get it! let me try again For Reaction on B: Set A as pivot, then total 3 moments (W CG and Wind) W(1.5) + CG(7) + Wind(13) = Moment WCG then to find RXN on B Force B = Moment WCG/8.5 = 4559 lb am I right?
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    Solving Billboard Problem with Equations of Motion

    hmm.. ok let me try again :D For Reaction on Point B: I'll set A as pivot, then total the Moment of W and CG. W(1.5) + CG(7) = Moment WCG then I'll divide it by 8.5 to get Reaction on point B? then B = 2724 lb so I will ignore wind load completely on calculating Rxn on B? or I will use...
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    Solving Billboard Problem with Equations of Motion

    The problem is... I don't know how should I get Rxn on point B... but anyways, I tried it using this calculations.. First I set point A is the pivot. Then I totaled the 3 moments (Weight, CG, and wind). W(1.5) + CG(7) + Wind(21.5) = Moment A Then I divided it by 8.5 (the distance from A to...
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    Solving Billboard Problem with Equations of Motion

    sorry for the late reply, been busy... ok here's what I did. For Reaction on point A: I replaced the 2 forces (W and CG) with a single force positioned at point A using this equation.. W(7) + CG(1.5) = F(8.5) F = 2,440 something. Then I reduce the Moment of that F by the Moment of the wind...
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    Solving Billboard Problem with Equations of Motion

    sorry but it's 11:30 pm here already.. got to sleep, have class tomorrow. Thanks for replying though, and thanks for correcting that 13 ft thingy. Ill try to post my calculations tomorrow.
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    Solving Billboard Problem with Equations of Motion

    Homework Statement The supporting structure of a billboard is attached to the ground by a pin at B, and its rear leg rests on the ground at A. Friction may be neglected. Point G is the center of gravity of the billboard and structure, which together weigh 2,800 lb. To prevent tipping over in...
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