Question

Two forces are applied at point B of beam AB. Determine the magnitude and
direction of their resultant. Figure:1

Answer 1

Explanation:

this is ans i hope mate ok

Answer 2

The magnitude and the direction of their resultant is 5Nm along O

Explanation:

The resultant moment of force is given by:

FORMULA             $\tau= r\times F=rFsin\theta$ where,  

r is the perpendicular distance between the point and the force line.

About A,

                     $τ(A) = r(1)F(1) + r(2)F(2)\\=0*10+50*10^{-2}*10=5Nm$

About B,

                    $τ(B)=r(2)F(2) + r(1)F(1)\\\\=0+10*50*10^{-2}=5Nm$

About the point O:

                   $\tau(O)=r(1)F(1)+r(2)F(2)\\=25\times 10^{-2}\times 10 + 25\times 10^{-2}\times 10=5Nm$