Question

Module 2 question 16

Answer 1

a ) The tension in a vertical strand = 68.6 * $10^{-5}$ N

b ) The tension in a horizontal strand = 149.13 * $10^{-5}$ N

c ) Tension in horizontal strand / Tension in vertical strand = 2.17

a ) The tension in a vertical strand,

m = 7 * $10^{-5}$ kg

g = 9.8 m / s²

Since the strand is vertical and the spider is motionless,

∑ $F_{y}$ = 0

T - mg = 0

T = 7 * $10^{-5}$ * 9.8

T = 68.6 * $10^{-5}$ N

b ) The tension in a horizontal strand,

θ = 13.0°

The strands left and right have same magnitude and makes the same angle with the horizontal. Resolving,

sin θ = $T_{y}$ / T

$T_{y}$ = T sin 13.0°

∑ $F_{y}$ = 0

$T_{y}$ + $T_{y}$ - mg = 0

T sin 13.0° + T sin 13.0° = 7 * $10^{-5}$ * 9.8

2 T sin 13.0° = 68.6 * $10^{-5}$

T = 68.6 * $10^{-5}$ / 0.46

T = 149.13 * $10^{-5}$ N

c ) Ratio of strands,

Tension in horizontal strand / Tension in vertical strand = 149.13 * $10^{-5}$ / 68.6 * $10^{-5}$

Tension in horizontal strand / Tension in vertical strand = 2.17

Therefore,

a ) The tension in a vertical strand = 68.6 * $10^{-5}$ N

b ) The tension in a horizontal strand = 149.13 * $10^{-5}$ N

c ) Tension in horizontal strand / Tension in vertical strand = 2.17

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