A block of mass m is connected to a peg at point D through a light and inextensible string of length 18 m. The length of section AB is 3 m. The point C of string is pulled
down by applying a force F of 128 N. Block A remains stationary. If the mass of block A is $\frac{16}{\sqrt{n}}$ kg, the value of n is

Question

Infinity Learn · Accepted Answer

From the figure, $$EC=(7.5)2$$−$$(4.5)2=6m$$ $$sin$$⁡θ$$=4.57.5=35$$⇒$$cos$$⁡θ$$=6.07.5=45$$ $$2F$$′$$cos$$⁡θ$$=128$$ ⇒ $$2F$$′×$$45=128$$⇒F′$$=128$$×$$58=80N$$ For block at A,F′$$=mgsin$$⁡$$45$$∘$$80=m$$×$$10$$×$$12$$⇒$$m=82kg$$∴ $$16n=82$$ Or  $$n=1682=2$$∴ $$n=2$$

Equilibrium of a particle

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Equilibrium of a particle

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A block of mass m is connected to a peg at point D through a light and inextensible string of length 18 m. The length of section AB is 3 m. The point C of string is pulleddown by applying a force F of 128 N. Block A remains stationary. If the mass of block A is 16n kg, the value of n is

From the figure, EC=(7.5)2−(4.5)2=6m sin⁡θ=4.57.5=35⇒cos⁡θ=6.07.5=45 2F′cos⁡θ=128 ⇒ 2F′×45=128⇒F′=128×58=80N For block at A,F′=mgsin⁡45∘80=m×10×12⇒m=82kg∴ 16n=82 Or n=1682=2∴ n=2

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