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.