When forces `F_1`, `F_2`, `F_3`, are acting on a particle of mass m such that `F_2` and `F_3` are mutually perpendicular, then the particle remains stationary. If the force `F_1` is now removed then the acceleration of the particle is

Options

A. `(R_3)/(m)`
B. `(R_(1)+R_(2))/(m)`
C. `(R_(1)-R_(2))/(m)`
D. `(R_1)/(m)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the particle and apply Newton's laws of motion.

### Step 1: Understand the Forces Acting on the Particle
We have three forces acting on a particle of mass \( m \):
- \( F_1 \)
- \( F_2 \)
- \( F_3 \)

Given that \( F_2 \) and \( F_3 \) are mutually perpendicular, we can visualize them as acting along the x-axis and y-axis respectively.

### Step 2: Condition for the Particle to be Stationary
For the particle to remain stationary, the net force acting on it must be zero. This means:
\[ F_1 + F_2 + F_3 = 0 \]
Since \( F_2 \) and \( F_3 \) are perpendicular, we can express the net force in terms of their magnitudes:
\[ F_1 = \sqrt{F_2^2 + F_3^2} \]

### Step 3: Removing Force \( F_1 \)
Now, if we remove the force \( F_1 \), the remaining forces acting on the particle are \( F_2 \) and \( F_3 \). Since \( F_2 \) and \( F_3 \) are still present, we need to find the resultant force acting on the particle.

### Step 4: Calculate the Resultant Force
The resultant force \( F_{net} \) when \( F_1 \) is removed is simply the vector sum of \( F_2 \) and \( F_3 \):
\[ F_{net} = \sqrt{F_2^2 + F_3^2} \]

### Step 5: Apply Newton's Second Law
According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
\[ F_{net} = m \cdot a \]
Substituting the expression for \( F_{net} \):
\[ \sqrt{F_2^2 + F_3^2} = m \cdot a \]

### Step 6: Solve for Acceleration \( a \)
To find the acceleration \( a \), we rearrange the equation:
\[ a = \frac{\sqrt{F_2^2 + F_3^2}}{m} \]

### Final Answer
Thus, the acceleration of the particle after removing force \( F_1 \) is:
\[ a = \frac{\sqrt{F_2^2 + F_3^2}}{m} \]

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