Physics, asked by gurnooraujla, 9 months ago

a mass of 5 kg is suspended by a rope of length 2m from a ceiling. a force of 50N in the horizontal direction is applied at the point P of the rope.
(i) what is the angle the rope makes with the vertical in equilibrium?
(ii) what is the tension in the rope? ​

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Answers

Answered by sreeh123flyback
11

Explanation:

for the 5kg mass

it is in equilibrium

so

T2=mg

T2=50N

at point P

T1cosø=T2 .........(1 ] [ since in equilibrium

T1sinø=50N.........(2]['xxxx'x]

T1sinø/T1cosø=50/50

tanø=1

ø=tan^-1(1)

Ø=45°

T2=50N

T1sin45=50

T1/_/2=50

T1=50root2N

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