In a right angle ABC, A=90^ and sides a,b,c are respectively 5 cm, 4 cm and 3 cm.

Question

In a right angle $	riangle ABC$, $\angle A=90^{\circ}$ and sides $a,b,c$ are respectively 5 cm, 4 cm and 3 cm. If a force $\vec F$ has moments $0,9$ and $16$ N\cdot cm units respectively about vertices $A,B$ and $C$, then the magnitude of $\vec F$ is:

Accepted Answer

Since the moment about $A$ is zero, the line of action of $\vec F$ passes through $A$. Taking $A$ as origin with line $y=mx$, the moments about $B$ and $C$ give:

$$\frac{3|\vec F|}{\sqrt{1+m^2}}=9,\qquad \frac{4|\vec F|}{\sqrt{1+m^2}}=16$$

Dividing: $m=\dfrac34$, and solving gives $|\vec F|=5$ N.

Solutions · 1