# Question #c16a5

Nov 21, 2016 To find out the minimum value of $l \text{ ie "l_"min}$ so that the woman can walk towards the other end W of the beam without tipping,we can say, when the woman reaches at the other end the moment of the weight of the beam (Mg) at G, the mid point of the beam and the moment of the weight of the woman (mg) at W about 2nd pivot at O are equal for $l = {l}_{\text{min}}$

Hence we can write

$O W \times m g = O G \times M g$

$\implies \left(P W - P O\right) \times m g = \left(P O - P G\right) \times M g$

$\implies \left(L - {l}_{\text{min}}\right) m g = \left({l}_{\min} - \frac{L}{2}\right) M g$

$\implies \left(L - {l}_{\text{min")/(l_"min}} - \frac{L}{2}\right) = \frac{M}{m}$

$\implies \left(L - {l}_{\text{min")/(l_"min}} - \frac{L}{2}\right) + 1 = \frac{M}{m} + 1$

$\implies \frac{\frac{L}{2}}{{l}_{\text{min}} - \frac{L}{2}} = \frac{M + m}{m}$

$\implies \left(\frac{L}{2}\right) = \left({l}_{\text{min}} - \frac{L}{2}\right) \times \frac{M + m}{m}$

$\implies \left({l}_{\text{min}} - \frac{L}{2}\right) = \frac{L}{2} \times \frac{m}{M + m}$

$\implies {l}_{\text{min}} = \frac{L}{2} \times \frac{m}{M + m} + \frac{L}{2}$

$\implies {l}_{\text{min}} = \frac{L}{2} \left(\frac{m}{M + m} + 1\right)$

$\implies {l}_{\text{min}} = \left(\frac{2 m + M}{2 M + 2 m}\right) \cdot L$

Nov 25, 2016

${l}_{\min} = \frac{1}{2} \left(\frac{M + 2 m}{M + m}\right) L$

#### Explanation:

The equilibrium limit is attained when the reaction support at the left most pivot is null. So Calling ${N}_{2}$ the second pivot reaction we have:

${N}_{2} - M g - m g = 0$

The moment regarding the left most pivot is

${N}_{2} {l}_{\min} - M g \frac{L}{2} - m g L = 0$

solving for ${N}_{2} , {l}_{\min}$ we obtain

${N}_{2} = \left(M + m\right) g$ and
${l}_{\min} = \frac{1}{2} \left(\frac{M + 2 m}{M + m}\right) L$