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*3-104. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location \((x, y)\) on the slab. Take $F_1 = 7 \text{ kN}$ and $F_2 = 9 \text{ kN}$. 3-105. The building slab is subjected to four parallel column loadings. Determine $F_1$ and $F_2$ if the resultant force acts through point \((17 \text{m}, 12 \text{m})\).

          *3-104. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location \((x, y)\) on the slab. Take $F_1 = 7 \text{ kN}$ and $F_2 = 9 \text{ kN}$.
3-105. The building slab is subjected to four parallel column loadings. Determine $F_1$ and $F_2$ if the resultant force acts through point \((17 \text{m}, 12 \text{m})\).
        
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*3-104. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Take F1 = 7  kN and F2 = 9  kN.
3-105. The building slab is subjected to four parallel column loadings. Determine F1 and F2 if the resultant force acts through point (17 m, 12 m).

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Can you please answer both questions? They both use the same image. Thank you. 3-104. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (xy) on the slab. Take Fi = 7 kN and F = 9 kN. 3-105. The building slab is subjected to four parallel column loadings. Determine F and I if the resultant force acts through point (17m, 12m). 9 kN F F2 6 kN x qmy 15m 19m 7m
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Transcript

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00:01 Firstly, we will write the given i .e.
00:03 F1 is 8kn, f2 is 9kn, f3 is 12kn and f4 is 6kn.
00:19 So let the resultant force f be.
00:28 So, applying equilibrium condition about x -axis or along x -direction, we will get summation of fx equals to 0.
00:40 So this will be fr -f1 -f2 -f3 -f4 equals to 0.
00:49 So from here we will get fr equals to 35kn.
00:57 So for location on the slab, we will apply the equilibrium condition for moments along x -direction i .e.
01:06 Summation of mx equals to 0.
01:10 So this will be fr multiplied by x -axis minus f1 multiplied by 0, f2 multiplied by 20 minus f3 multiplied by 8 minus f4 multiplied by 20 equals to 0.
01:26 So from here we will get fr multiplied by x minus 180 minus 96 minus 120 equals to 0...
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