If $$ \vec{F}=2 i+3 j $$ acts at point $$ C $$ and $$ \overline{A B}=4 i+6 j $$ and $$ \overline{M_{A}}=\left(m^{2}+4 ight) k, \overline{M_{B}}= $$ $$ (4 m) k $$, then $$ m=\cdots \cdots $$ $$ \begin{array}{llll} ext { (a) } 3 & ext { (b) } 2 & ext { (c) } 4 & ext { (d) } 1\end{array} $$

Question

If $$ \vec{F}=2 i+3 j $$ acts at point $$ C $$ and $$ \overline{A B}=4 i+6 j $$ and $$ \overline{M_{A}}=\left(m^{2}+4ight) k, \overline{M_{B}}= $$ $$ (4 m) k $$, then $$ m=\cdots \cdots $$ $$ \begin{array}{llll}	ext { (a) } 3 & 	ext { (b) } 2 & 	ext { (c) } 4 & 	ext { (d) } 1\end{array} $$

UpStudy AI · Accepted Answer

To determine the value of $$ m $$, let's analyze the given information step by step.

1. **Given Vectors:**
   - Force vector: $$ \vec{F} = 2\mathbf{i} + 3\mathbf{j} $$ acting at point $$ C $$.
   - Vector $$ \overline{AB} = 4\mathbf{i} + 6\mathbf{j} $$.

2. **Moments:**
   - Moment about point $$ A $$: $$ M_A = (m^2 + 4)\mathbf{k} $$.
   - Moment about point $$ B $$: $$ M_B = 4m\mathbf{k} $$.

3. **Equilibrium Condition:**
   For equilibrium, the total moment about any point should be zero. This implies:
   $$
   M_A = M_B
   $$
   Substituting the given moments:
   $$
   m^2 + 4 = 4m
   $$
   Rearranging the equation:
   $$
   m^2 - 4m + 4 = 0
   $$
   
4. **Solving the Quadratic Equation:**
   $$
   m = \frac{4 \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2
   $$
   
   So, $$ m = 2 $$.

**Answer:** Option (b) $$ 2 $$.
$$ m = 2 $$

If acts at point and and
, then

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Physics Questions

If acts at point and and , then