Determine the sum of the moments of the three loads about A

Determine the sum of the moments of the three loads about A. 1 kN-m 6 kN-m 11 kN-m 25 kN-m 47 kN-m none of the above

Solution

Direction of the force 5 kN

tan^-1(3/4) = 36.87^o with vetical axis

So vertical component of the force is 5 cos 36.87^o = 3.99999 kN= 4 kN

And , horizontal component of the force 5 sin 36.87^o = 3 kN (towards right)

Now clockwise moments: ( Due to force 6 kN and vertical & horizontal component of the force 5 kN i.e. 4 kN & 3 kN)

6 kN. x 0.5 m + 4 kN x (1.5 + 0.5)m + 3 kN x(3+0.5+0.5)m = 23 kN-m

anti-clockwise moment (Due to force 8 kN)

8 kN x 3 m = -24 kN-m ( anti clockwise moment is negative)

So sum of moments

23 kN-m - 24 kN-m = -1 kN-m................................(Ans)