First slide

Work

Question

A mass of M kg is suspended by a weight less string the horizontal force that requires to displace it until the string making an angle 45° with the initial vertical direction is

Moderate

A

$M g (\sqrt{2} - 1)$
B

$M g (\sqrt{2} + 1)$
C

$M g \sqrt{2}$
D

$\frac{M g}{\sqrt{2}}$

Solution

$w o r k d o n e b y h o r i z o n t a l f o r c e i s W = F \times ℓ \sin 45 °$
$W o r k d o n e b y h o r i z o n t a l f o r c e = \frac{F ℓ}{\sqrt{2}}$ ---------(1)
   $g a i n i n P E o f p e n d u l u m = M g h P E = M g (ℓ - ℓ \cos 45 °) P E = M g ℓ (1 - \frac{1}{\sqrt{2}})$
$P E = M g ℓ (\frac{\sqrt{2} - 1}{\sqrt{2}})$ -----------(2)
   $s u b s t i t u t e e q u a t i o n (1) i n (2) ∴ \frac{F ℓ}{\sqrt{2}} = M g ℓ (\frac{\sqrt{2} - 1}{\sqrt{2}})$
   $\Rightarrow F = M g (\sqrt{2} - 1)$

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