Question

The forces of $60\text{ N}$ and $80\text{ N}$ are in equilibrium at a point, acting at an angle of $90^\circ$ to each other. Find the magnitude of the third force?

• A. $100\text{ N}$
• B. $120\text{ N}$
• C. $140\text{ N}$
• D. $160\text{ N}$

Hint:

This is a standard $(3, 4, 5)$ right-triangle ratio scaled by a factor of $20$:
$3(20) = 60\text{ N}$
$4(20) = 80\text{ N}$
$5(20) = 100\text{ N}$ (Resultant/Third force magnitude)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the magnitude of a third force required to maintain static equilibrium at a point alongside two perpendicular forces of $60\text{ N}$ and $80\text{ N}$.

Step 2: Key Formula or Approach:
For three forces $\mathbf{F_1}$, $\mathbf{F_2}$, and $\mathbf{F_3}$ to be in equilibrium:
\[ \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} = 0 \implies \mathbf{F_3} = -(\mathbf{F_1} + \mathbf{F_2}) \]
This means the third force must be equal in magnitude and opposite in direction to the resultant of the first two forces ($R$):
\[ F_3 = R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta} \]

Step 3: Detailed Explanation:

• Given values:
$F_1 = 60\text{ N}$
$F_2 = 80\text{ N}$
$\theta = 90^\circ$ (angle between $F_1$ and $F_2$)

• Since the forces are perpendicular, $\cos(90^\circ) = 0$.

• The magnitude of the resultant force $R$ is:
\[ R = \sqrt{F_1^2 + F_2^2} \]
\[ R = \sqrt{60^2 + 80^2} \]
\[ R = \sqrt{3600 + 6400} \]
\[ R = \sqrt{10000} = 100\text{ N} \]

• To balance this resultant force of $100\text{ N}$ and maintain static equilibrium at the point, the third force must have a magnitude of exactly $100\text{ N}$ and act in the opposite direction.

Step 4: Final Answer:
The magnitude of the third force is $100\text{ N}$.