Question

one end of a string 0.5 m long is fixed to a point A and the other end is fastened to a small object of weight 8 N.the object is pulled aside by a horizontal force F,until it is 0.3 m from the vertical through A.find the magnitude of the tension T in the string and the force F.

Answer 1

Answer:

Explanation:

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Answer 2

Given :-

AC = 0.5 m

BC = 0.3 m

AB = 0.4 m

CosØ = AB/AC = 4/5

SinØ = BC/AC = 3/5

Using Lami's theorem,

F/Sin(180°-Ø) = 8/Sin(90°+Ø) = T/Sin90°

T = 8/CosØ (Sin(90°+Ø) = CosØ).

T = 8 × 5/4

T = 10N.

Again,

F×5/3 = 8 × 5/4

F = 24/4

F = 6 N

Hence,

Tension = T = 10N

Force = F = 6N