ABCD is a right-angled trapezium at B in which $$ \overrightarrow{A D} / / \overrightarrow{B C}, \mathrm{AB}=8 \mathrm{~cm} $$, $$ \mathrm{BC}=15 \mathrm{~cm} $$ and $$ \mathrm{AD}=9 \mathrm{~cm} $$. Forces of magnitudes $$ \mathrm{F}, 44 $$ and $$ 68 \mathrm{gm} . \mathrm{wt} $$ act in $$ \overrightarrow{D A}, \overrightarrow{D C} $$ and $$ \overrightarrow{A C} $$ respectively. If the line of action of the resultant passes through B , Then the value of $$ \mathrm{F}=\ldots \ldots \ldots $$ gm.wt. $$ \begin{array}{llll} ext { (a) } 114 & ext { (b) } 126 & ext { (c) } 156 & ext { (d) } 184\end{array} $$

Question

ABCD is a right-angled trapezium at B in which $$ \overrightarrow{A D} / / \overrightarrow{B C}, \mathrm{AB}=8 \mathrm{~cm} $$, $$ \mathrm{BC}=15 \mathrm{~cm} $$ and $$ \mathrm{AD}=9 \mathrm{~cm} $$. Forces of magnitudes $$ \mathrm{F}, 44 $$ and $$ 68 \mathrm{gm} . \mathrm{wt} $$ act in $$ \overrightarrow{D A}, \overrightarrow{D C} $$ and $$ \overrightarrow{A C} $$ respectively. If the line of action of the resultant passes through B , Then the value of $$ \mathrm{F}=\ldots \ldots \ldots $$ gm.wt. $$ \begin{array}{llll}	ext { (a) } 114 & 	ext { (b) } 126 & 	ext { (c) } 156 & 	ext { (d) } 184\end{array} $$

UpStudy AI · Accepted Answer

To solve the problem, let's analyze the given right-angled trapezium $$ ABCD $$ with the right angle at $$ B $$.

### Given:
- $$ AB = 8 $$ cm (vertical side)
- $$ BC = 15 $$ cm (horizontal side)
- $$ AD = 9 $$ cm (parallel to $$ BC $$)
- Forces:
  - $$ F $$ acts along $$ DA $$
  - $$ 44 $$ gm.wt acts along $$ DC $$
  - $$ 68 $$ gm.wt acts along $$ AC $$
- The resultant force passes through $$ B $$

### Step-by-Step Solution:

1. **Coordinate System Setup:**
   - Let $$ B $$ be at the origin $$ (0,0) $$.
   - $$ A $$ is at $$ (0,8) $$.
   - $$ C $$ is at $$ (15,0) $$.
   - $$ D $$, being parallel to $$ BC $$, is at $$ (9,8) $$.

2. **Vector Representation of Forces:**
   - **Force $$ F $$ along $$ DA $$:**
     - Direction of $$ DA $$ is from $$ D $$ to $$ A $$, i.e., $$ (-9, 0) $$.
     - Force vector: $$ \vec{F}_{DA} = F \cdot (-1, 0) $$.
   
   - **Force $$ 44 $$ gm.wt along $$ DC $$:**
     - Direction of $$ DC $$ is from $$ D $$ to $$ C $$, i.e., $$ (6, -8) $$.
     - Unit vector: $$ \frac{(6,-8)}{10} $$.
     - Force vector: $$ \vec{F}_{DC} = 44 \cdot \left(\frac{6}{10}, \frac{-8}{10}\right) = (26.4, -35.2) $$.

- **Force $$ 68 $$ gm.wt along $$ AC $$:**
     - Direction of $$ AC $$ is from $$ A $$ to $$ C $$, i.e., $$ (15, -8) $$.
     - Unit vector: $$ \frac{(15,-8)}{17} $$.
     - Force vector: $$ \vec{F}_{AC} = 68 \cdot \left(\frac{15}{17}, \frac{-8}{17}\right) = (60, -32) $$.

3. **Condition for Resultant Passing Through $$ B $$:**
   - The sum of moments about point $$ B $$ must be zero:
     $$
     \text{Moment}_{DA} + \text{Moment}_{DC} + \text{Moment}_{AC} = 0
     $$
   - Calculating each moment:
     - $$ \text{Moment}_{DA} = (9 \cdot 0) - (8 \cdot (-F)) = 8F $$
     - $$ \text{Moment}_{DC} = (9 \cdot (-35.2)) - (8 \cdot 26.4) = -316.8 - 211.2 = -528 $$
     - $$ \text{Moment}_{AC} = (0 \cdot (-32)) - (8 \cdot 60) = -480 $$
   - Summing moments:
     $$
     8F - 528 - 480 = 0 \implies 8F = 1008 \implies F = 126 \text{ gm.wt}
     $$

### **Answer:**
**Option (b) 126 gm.wt.**
The value of $$ F $$ is 126 gm.wt.

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