Question:hard

Two unlike parallel forces $2\,N$ and $16\,N$ act at the ends of a uniform rod of length $21\,cm$. The point where the resultant acts is at a distance of ____ from the greater force.

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For unlike parallel forces, the resultant always lies outside the two forces and closer to the larger force.
Updated On: Jun 5, 2026
  • $4\,cm$
  • $3\,cm$
  • $2\,cm$
  • $1\,cm$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understand unlike parallel forces.
Two parallel forces that point in opposite directions are called unlike forces. For such a pair, the single resultant force acts outside both of them, on the side of the bigger force. This is the picture we keep in mind.

Step 2: Use the moment rule.
To locate the resultant we balance turning effects, called moments. A moment is force times its distance. For the resultant point, the moments of the two given forces must balance: $F_1 d_1=F_2 d_2$.

Step 3: List the data.
The forces are $F_1=2\,\text{N}$ and $F_2=16\,\text{N}$, sitting at the ends of a rod $21\,\text{cm}$ long. Let the resultant be at distance $x$ from the larger $16\,\text{N}$ force. Then it is $21+x$ from the smaller force, since it lies outside on the big force side.

Step 4: Write the moment balance.
Set the moment of the big force equal to the moment of the small force about the resultant point. \[ 16x=2\,(21+x) \]

Step 5: Simplify.
Open the right side. \[ 16x=42+2x \] Bring the $x$ terms together: $16x-2x=42$, so $14x=42$.

Step 6: Solve for $x$.
Divide both sides by $14$. \[ x=\frac{42}{14}=3\,\text{cm} \] So the resultant acts $3\,\text{cm}$ from the greater force, which is option 2. \[ \boxed{3\ \text{cm}} \]
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