Question

Three ropes A, B and C are tied together in one single knot K.

Answer 1

The tension in the rope B is determined as 10.9 N.

Vertical angle of cable B

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

Angle between B and C

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

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Answer 2

Hello!

This is an example of a force summation in the vertical direction.

We have the tension of rope A upward (+), and the equal vertical components of the tensions of rope B and C downward (-).

These forces sum to zero, since the knot is stationary.

$\Sigma F = T_A - T_{By} - T_{Cy} \\\\0 = T_A - T_{By} - T_{Cy}$

Ropes 'B' and 'C' form equivalent angles from the vertical. (If you were to draw a line from rope A down). We can use right-triangle trig to determine the angle:

$tan^{-1}(\frac{O}{A}) = \theta$

The ropes are 5 m long and 2 m tall, which are the opposite and adjacent sides respectively:
$tan^{-1}(\frac{5}{2}) = 68.2^o$

The vertical components are the adjacent sides from this angle, so, we would use cosine.

$0 = T_A - T_Bcos\theta - T_Ccos\theta$

Rope 'B' and 'C' have the same tensions since they form the same angle with the vertical and are the same length, so we can call them 'T'.

$0 = T_A - 2Tcos\theta$

Solving for 'T':

$2Tcos\theta = T_A \\\\T = \frac{T_A}{2cos\theta}\\\\T = \frac{65.3}{2cos(68.2)} = \boxed{87.92 N}$