# Statics equilibrium?

Mar 24, 2017

$M g$

#### Explanation:

Calling

${f}_{s} = \left(\frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right) {t}_{s}$
${f}_{h} = \left(- 1 , 0\right) {t}_{h}$
$p = \left(0 , - M g\right)$

the equilibrium equation is

${f}_{s} + {f}_{h} + p = \left(0 , 0\right)$

or

$\left\{\begin{matrix}\frac{\sqrt{2}}{2} {t}_{s} - {t}_{h} = 0 \\ \frac{\sqrt{2}}{2} {t}_{s} - M g = 0\end{matrix}\right.$

solving for ${t}_{s} , {t}_{h}$ we obtain

${t}_{s} = \sqrt{2} M g$ and
${t}_{h} = M g$

Mar 25, 2017

$F = M g$

#### Explanation:

$\tan \beta = \frac{M g}{F}$

$F = \frac{M g}{\tan} \beta$

$\tan {45}^{o} = 1$

$F = \frac{M g}{1}$

$F = M g$

Mar 29, 2017

Option (1)

#### Explanation:

Let $A B$ be initial position of mass $M$ and $A B '$ be its final position.

We see that the mass has moved a horizontal distance $= C B '$
and consequently it has moved through a vertical distance of $B C$
Due to movement in the vertical direction it has gained
Gravitational potential energy$= M g | \vec{B C} |$ ......(1)
Due to Law of conservation of energy this must be equal to work done by an effective Horizontal force $\vec{F}$ through a distance $\vec{C B '}$

I have used effective as position of the force keeps on changing as the mass moves up.
Work done by the force$= \vec{F} \cdot \vec{C B '}$

As shown in the fig above, angle between the two is ${0}^{\circ}$
$\implies$ Work done by the force$= | \vec{F} | | \vec{C B '} |$ .......(2)

Equating (1) and (2) we get
$M g \overline{B C} = | \vec{F} | \overline{C B '}$
$\implies | \vec{F} | = \frac{M g \overline{B C}}{\overline{C B '}}$ ......(3)

Let $L$ be length of the string. In $\Delta A C B '$
$\overline{C B '} = L \sin {45}^{\circ} = \frac{L}{\sqrt{2}}$
and $\overline{B C} = L - \frac{L}{\sqrt{2}}$ ($\Delta A C B '$ is an isosceles triangle)

Inserting these values in (3) we get
$| \vec{F} | = \left(M g\right) \frac{L - \frac{L}{\sqrt{2}}}{\frac{L}{\sqrt{2}}}$
$\implies | \vec{F} | = M g \left(\sqrt{2} - 1\right)$