# Following sets of forces act on a body. In which case the resultant cannot be zero.

$\begin{array}{1 1}(A)\;10N,10N,10N \\(B)\;10N,10N,20N \\(C)\;10N,20N,20N\\(D)\;10N,20N,40N \end{array}$

In a triangle of sides, a,b and c
$a+b > c$ so $a > (c-b)$
$b+c > a$ so $b > (a-c)$
$c+a >b$ so $c > (b-a)$
Since three forces in equilibrium can be represented as the three sides of a triangle and in a triangle each side (representing a force) is numerically greater than the different of the other two sides (forces), hence for equilibrium each force should be greater than or equal to the difference of the other two.
This is satisfied by (a) ,(b) and (c) and not by (d)
Hence D is the correct answer.