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Pachacha [2.7K]

3 years ago

11

A strong weightless rope has a mass, m, hanging from the middle of it. The tension force on each rope is 25 N, and the rope droo

ps at an angle of 20.0 degrees. How much mass is hanging from the rope?

1 answer:

bezimeni [28]3 years ago

4 0

By using Lami's theorem, Mass m = 1.75 kg approximately

Given that a strong weightless rope has a mass, m, hanging from the middle of it. If the tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees to the horizontal.

By using Lami's theorem, we can get how much mass is hanging from the rope.

Let the angle between the rope = α = 180 - 40

α = 140 degrees

The angle between one of the rope and mass = β = 20 + 90

β = 110 degrees

The angle between the mass and the other rope = γ = 360 - (140 + 110)

γ = 360 - 250

γ = 110 degrees

W/ sinα = T/ sinβ = T/sinγ

W/ sinα = T/ sinβ

Substitute all the necessary parameters

W/sin140 = 25/sin 110

W / 0.643 = 25 / 0.939

W = 17.1 N

Weight W = mg

17.1 = 9.8m

mass m = 17.1/9.8

Mass m = 1.7455 kg

Mass m = 1.75 kg approximately

Therefore, 1.75 kg mass is hanging from the rope.

Learn more about resolution of forces here: brainly.com/question/1858958

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Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

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magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

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<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

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$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

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$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

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$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

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