Question

Three forces 3 N, 4 N, and 5 N act at a point. For equilibrium, angle between 3 N and 4 N is

• A. $60^\circ$
• B. $90^\circ$
• C. $120^\circ$
• D. $180^\circ$

Hint:

This problem is based on the Pythagorean triple $(3, 4, 5)$.
For any forces that form a right-angled triangle, the angle between the two base forces is always $90^\circ$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the angle between the $3\text{ N}$ and $4\text{ N}$ forces such that the system of three forces ($3\text{ N}$, $4\text{ N}$, and $5\text{ N}$) acting at a point is in static equilibrium.

Step 2: Key Formula or Approach:
For a particle to be in static equilibrium under three coplanar forces, the vector sum of the forces must be zero:
\[ \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} = 0 \implies \mathbf{F_1} + \mathbf{F_2} = -\mathbf{F_3} \]
Taking the magnitude of both sides:
\[ F_1^2 + F_2^2 + 2F_1F_2 \cos\theta = F_3^2 \]
where $\theta$ is the angle between $\mathbf{F_1}$ and $\mathbf{F_2}$.

Step 3: Detailed Explanation:

• Let $F_1 = 3\text{ N}$, $F_2 = 4\text{ N}$, and $F_3 = 5\text{ N}$.

• Substitute these values into the vector equilibrium equation:
\[ 3^2 + 4^2 + 2(3)(4) \cos\theta = 5^2 \]
\[ 9 + 16 + 24 \cos\theta = 25 \]
\[ 25 + 24 \cos\theta = 25 \]
\[ 24 \cos\theta = 0 \]
\[ \cos\theta = 0 \implies \theta = 90^\circ \]

• This shows that the $3\text{ N}$ and $4\text{ N}$ forces must act perpendicular to each other, allowing their resultant ($\sqrt{3^2 + 4^2} = 5\text{ N}$) to be perfectly balanced by the opposing $5\text{ N}$ force.

Step 4: Final Answer:
The angle between the $3\text{ N}$ and $4\text{ N}$ forces must be $90^\circ$.