Question

Find F for given figure

a. $\dfrac{Mg}{\sqrt{2}}$
b. $Mg\sqrt{2}$
c. $Mg$
d. $3Mg$

Answer 1

Hint : For static equilibrium to be maintains on a point, following conditions are must
Net force along the x-axis $\Sigma {{F}_{x}}=0$
Net force along the y-axis $\Sigma {{F}_{y}}=0$
Net force along the z-axis $\Sigma {{F}_{z}}=0$
Net moment about all the three axes also should be zero.

Complete Step By Step Answer:
For point P to be in equilibrium all force along x and y-axis must be balanced.
Let assume tension in PQ is T

Here is the F.B.O of point P. For equilibrium at point P.

$\Sigma {{F}_{x}}=0$
$T\cos {{45}^{\circ }}-F=0.........\left( i \right)$
$\Sigma {{F}_{y}}=0$
$T\sin {{45}^{\circ }}-mg=0.............\left( ii \right)$
From equation (i) and (ii)
$T=\sqrt{2}mg$
$F=T\cos {{45}^{\circ }}=\sqrt{2}mg\left( \dfrac{1}{\sqrt{2}} \right)$
$F=mg$
Hence for a given figure $F=mg$ .

Note :
Students should resolve the con point of force along the x and y-axis direction very carefully. And here tension in PQ will not be equal to mg. Though it is a single rope. Tension in one rope is some if there is a press less pulley at point P. But because at point force F is applied therefore tension will not be the same throughout the rope by which block is hanging.