The building slab is subjected to four parallel column loadi

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=30kN, F2=40kN

Solution

First we find the resultant force on the body

Since , F1 = -30 k [ negative sign is used because it is opposite direction of positive z axis ]

F2 = -40KN k

F3 =-50KN k

F4 = -20KN k

Let theresultant be F and act at x,y coordinate

Therefore using newton\'s law of motion, F + F1 +F2+F3+F4 =0

therefore, F =-(-20-50-30-40)k=140KN k

now F1 is acting at 0i+11j [ the coordinates are written in vector form so that their torque can be found easily]

F2 at 10i+13j , F3 at 4i+3j , F4 at 10i

Now taking torque of all these forces about O gives us -30k*11j -40k*(10i+13j) -50k*(4i+3j) -20k*10i + Fk(xi+yj)   

This moment should be zero, therefore, -330(-i) -400(j) -520(-i) -200(j) -150(-i) -200j +F*x j + F*y(-i) =0

Equating i and j terms to zero separately and putting F = 140 we have, y=1000/140m and x = 800/140m - Ans