A mass of 10kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is q = tan^-¹ (x * 10^-¹). The value of x is……………..

(Given, g = 10 m/s²)

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Payal Gupta | Contributor-Level 10

6 months ago

F.B.D. of point 'P'

$T s i n θ = 30$ - (1)

$T c o s θ = 100$ - (2)

$(1) \div (2)$

$t a n θ = \frac{3}{10} \Rightarrow θ = t a n^{- 1} (3 * 1 0^{- 1})$

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A mass of 10kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is q = tan-1 (x * 10-1). The value of x is…………….. (Given, g = 10 m/s2)