Question
$$O$$ is the centre of an equilateral $$\Delta ABC.$$ $${F_1},{F_2}$$ and $${F_3}$$ are three forces acting along the sides $$AB,\,BC$$ and $$AC$$ as shown in figure. What should be the magnitude of $${F_3},$$ so that the total torque about $$O$$ is zero?
$$O$$ is the centre of an equilateral $$\Delta ABC.$$ $${F_1},{F_2}$$ and $${F_3}$$ are three forces acting along the sides $$AB,\,BC$$ and $$AC$$ as shown in figure. What should be the magnitude of $${F_3},$$ so that the total torque about $$O$$ is zero?
A.
$$\frac{{\left( {{F_1} + {F_2}} \right)}}{2}$$
B.
$$\left( {{F_1} - {F_2}} \right)$$
C.
$$\left( {{F_1} + {F_2}} \right)$$
D.
$$2\left( {{F_1} + {F_2}} \right)$$
Answer :
$$\left( {{F_1} + {F_2}} \right)$$
Solution :
Let $$r$$ be the perpendicular distance of $${F_1},{F_2}$$ and $${F_3}$$ from $$O$$ as shown in figure.
The torque of force $${F_3}$$ about $$O$$ is clockwise, while torque due to $${F_1}$$ and $${F_2}$$ are anticlockwise.
For total torque to be zero about $$O,$$ we must have $${F_1}r + {F_2}r - {F_3}r = 0$$
$$ \Rightarrow {F_3} = {F_1} + {F_2}$$
Let $$r$$ be the perpendicular distance of $${F_1},{F_2}$$ and $${F_3}$$ from $$O$$ as shown in figure.
The torque of force $${F_3}$$ about $$O$$ is clockwise, while torque due to $${F_1}$$ and $${F_2}$$ are anticlockwise.
For total torque to be zero about $$O,$$ we must have $${F_1}r + {F_2}r - {F_3}r = 0$$
$$ \Rightarrow {F_3} = {F_1} + {F_2}$$
