Question
In a right angle $$\Delta ABC,\,\angle A = {90^ \circ }$$ and sides $$a,\,b,\,c$$ are respectively, $$5\,cm,$$ $$4\,cm$$ and $$3\,cm.$$ If a force $$\overrightarrow F $$ has moments $$0,\,9$$ and $$16$$ in $$N\,cm$$ units respectively about vertices $$A,\,B$$ and $$C,$$ then magnitude of $$\overrightarrow F $$ is :
A.
$$9$$
B.
$$4$$
C.
$$5$$
D.
$$3$$
Answer :
$$5$$
Solution :
Since, the moment about $$A$$ is zero, hence $$\overrightarrow F $$ passes through $$A.$$ Taking $$A$$ as origin.
Let the line of action of force $$\overrightarrow F $$ be $$y = mx.$$ (see figure)
Moment about $$B = \frac{{3m}}{{\sqrt {1 + {m^2}} }}\left| {\overrightarrow F } \right| = 9......\left( 1 \right)$$
Moment about $$C = \frac{4}{{\sqrt {1 + {m^2}} }}\left| {\overrightarrow F } \right| = 16......\left( 2 \right)$$
Dividing $$\left( 1 \right)$$ by $$\left( 2 \right),$$ we get :
$$m = \frac{3}{4} \Rightarrow \left| {\overrightarrow F } \right| = 5N.$$
Since, the moment about $$A$$ is zero, hence $$\overrightarrow F $$ passes through $$A.$$ Taking $$A$$ as origin.
Let the line of action of force $$\overrightarrow F $$ be $$y = mx.$$ (see figure)
Moment about $$B = \frac{{3m}}{{\sqrt {1 + {m^2}} }}\left| {\overrightarrow F } \right| = 9......\left( 1 \right)$$
Moment about $$C = \frac{4}{{\sqrt {1 + {m^2}} }}\left| {\overrightarrow F } \right| = 16......\left( 2 \right)$$
Dividing $$\left( 1 \right)$$ by $$\left( 2 \right),$$ we get :
$$m = \frac{3}{4} \Rightarrow \left| {\overrightarrow F } \right| = 5N.$$