51. Two forces act an angle of $$ 120^{\circ} $$. If the greater force is 50 kg and their resultant is perpendicular to the smaller force, the smaller force is A. 20 kg B. 25 kg C. 30 kg D. 35 kg

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To solve this problem, let's analyze the forces using vector components.

**Given:**
- Two forces act at an angle of $$ 120^{\circ} $$.
- The greater force $$ F_1 $$ is 50 kg.
- The resultant force $$ R $$ is perpendicular to the smaller force $$ F_2 $$.

**Objective:**
Find the magnitude of the smaller force $$ F_2 $$.

**Approach:**

1. **Vector Representation:**
   - Assume $$ F_2 $$ is along the x-axis: $$ F_2 = (x, 0) $$.
   - $$ F_1 $$ makes a $$ 120^{\circ} $$ angle with $$ F_2 $$:
     $$
     F_1 = (50 \cos 120^{\circ}, 50 \sin 120^{\circ}) = (-25, 25\sqrt{3})
     $$
   
2. **Resultant Force ($$ R $$):**
   $$
   R = F_1 + F_2 = (-25 + x, 25\sqrt{3})
   $$
   
3. **Perpendicular Condition:**
   - The resultant $$ R $$ is perpendicular to $$ F_2 $$.
   - For two vectors to be perpendicular, their dot product must be zero:
     $$
     R \cdot F_2 = (-25 + x)x + 25\sqrt{3} \cdot 0 = 0
     $$
     $$
     (-25 + x)x = 0
     $$
   
4. **Solving for $$ x $$:**
   $$
   x(x - 25) = 0
   $$
   $$
   x = 0 \quad \text{or} \quad x = 25
   $$
   - $$ x = 0 $$ is not a valid physical solution in this context.
   - Therefore, $$ x = 25 $$ kg.

**Answer:**  
**B. 25 kg**
The smaller force is 25 kg.

Two forces act an angle of . If the greater force is 50 kg and their
resultant is perpendicular to the smaller force, the smaller force is
A. 20 kg
B. 25 kg
C. 30 kg
D. 35 kg

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Two forces act an angle of . If the greater force is 50 kg and their resultant is perpendicular to the smaller force, the smaller force is A. 20 kg B. 25 kg C. 30 kg D. 35 kg