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3 years ago

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The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 9.1 kN. Determine the requ

ired tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.

1 answer:

Oksana_A [137]3 years ago

7 0

Answer:

T_ac = 6.586 KN

R = 10.51 KN

Explanation:

Given:

- Tension in cable T_ab = 9.1 KN

Find:

- Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A

- Determine the magnitude R of this downward force.

Solution:

- Compute the three angles as shown in figure attached, a, B , y:

a = arctan (40/50) = 38.36 degrees

B = arctan (50/30) = 59.04 degrees

y = 180 - 38.36 = 82.6 degrees

- Use cosine rule to calculate R and F_ac as follows:

sin(a) / T_ac = sin(B) / T_ab = sin(y) / R

sin(38.36) / T_ac = sin(59.04) / 9.1 = sin(82.06) / R

T_ac = 9.1 * ( sin(38.36) / sin(59.04) )

T_ac = 6.586 KN

R = 9.1 * ( sin(82.06) / sin(59.04) )

R = 10.51 KN

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3 years ago

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Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

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1 AU is defined as the distance between the earth and the sun.

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But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

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That can be expressed in units of days

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$v = \frac{2 \pi r}{T}$

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For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

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$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

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$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

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The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago