Recent content by bcoolbelal

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    Calculating Factor of Safety: Ultimate Tensile Strength and Shear for 7mm Bolt

    sory for this prob try this link http://i826.photobucket.com/albums/zz190/bcoolbelal/Untitled.jpg
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    Calculating Factor of Safety: Ultimate Tensile Strength and Shear for 7mm Bolt

    **factor of safety** Homework Statement Im having problems with this, any help would be much appreciated http://i826.photobucket.com/albums/zz190… The material for the bolt shown in the angled joint above has an ultimate tensile strength of 218 MPa and a shear of 230 MPa. The...
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    Calculating Resultant Force of a Noncurrent System

    No I don't have the equ for direction of R. So u mean I need a single point where i takey moments from for all 4. Oh it's starting to make more sense now thankx. I'll have a go at it and reply with the result thankx
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    Calculating Resultant Force of a Noncurrent System

    i think i got it now, caan u please comfirm that the following are corect and that all i need help with: The calculations for the 4kN M=F X D F Y = F Sin Ø = 4 Sin 45 = 2.82 F x = F Cos Ø = 4 Cos 45 = 2.82 ACM = F X D = 2.82 X 1 = 2.82 CM = F X D = 2.82 X 0 = 0 ∑ FX = 2.82 ∑ FY...
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    Calculating Resultant Force of a Noncurrent System

    typing errors sorted, so where to now, can u please guide me on the 5kn
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    Calculating Resultant Force of a Noncurrent System

    oh, the X axis is where the 6.93 comes from
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    Calculating Resultant Force of a Noncurrent System

    and i believe the way of finsing the resultant force is: R^2 = (∑ FX )^2 + (∑ Fy)^2 R = √((∑ FX) ^2 + ((∑ Fy)^2 is that equation right
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    Calculating Resultant Force of a Noncurrent System

    thankx for the explination however that was only a typing problem, anyway I am currently stuck on the 5kN fx and yx as it does not state an angle
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    Calculating Resultant Force of a Noncurrent System

    and i also had a go at the 3 kN one. abd it turnd out as the following: M= F X D F Y = F Sin Ø = 3 Sin 0 = 0 F x = F Sin Ø = 3Cos 0 = 3 ACM = F X D = 0 X 0.5 = 0 CM = F X D = 3 X 0 = ∑ FX = 3 ∑ FY = 0 ∑ M = 0...
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    Calculating Resultant Force of a Noncurrent System

    OK THANKX FOR THAT BIT, i re don it and came out with the following F Y = F Sin Ø = 8 Sin 30 = 4 F x = F Sin Ø = 8 Cos 150 = -6.93 ACM = F X D = 4 X 0.5 = 2 CM = F X D = -6.93X 0 = 0 sory i did not get ehat u ment after this bit
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    Calculating Resultant Force of a Noncurrent System

    well i had a go at it, and this iss how it went: F Y = F Sin Ø = 8 Sin -30 = -4 F x = F Sin Ø = 8 Cos -30 = 6.93 ACM = F X D = -4 X 0.5 = -2 CM = F X D = 6.93X 0 = 0 (In FY and FX i put -30 angle cause the angle is acting anti clock wise) so what to next
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    Calculating Resultant Force of a Noncurrent System

    the problem is i don't get the question, the first attempt is. the 4KN side. M=F X D F Y = F Sin Ø = 4 Sin 45 = 2.82 F x = F Sin Ø = 4 Cos 45 = 2.82 ACM = F X D = 2.82 X 1 = 2.82NA CM = F X D = 2.82 X 0 = 0 then i don't know where to go from there
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    Calculating Resultant Force of a Noncurrent System

    Homework Statement Calculate the magnitude, direction and sense of the resultant force of the noncurrent force system shown, determine where the resultant intersects the bottom of the shape with respect to Point A Homework Equations...
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