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During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

Answer :

Manetho

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

Answer:

Magnitude - 11.83 Degree

Direction - 422.42 N

Explanation:

Given data:

Downward force on wire 176 N

Angle made by left section of wire 12.5 degree with horizontal

Tension force = 413 N

From figure

Applying quilibrium principle at point A

The vertical and horizontal force is 0

then we have

[tex]Tcos\theta = 413 N[/tex]   ........1

[tex]176 = 413 sin 12.5 + Tsin\theta[/tex]     .......2

[tex]Tsin\theta = 176 - 89.39  = 86.6[/tex].......3

divide equation 3 by 1

we get

[tex]\theta = tan^{-1} (0.2096)[/tex]

[tex]theta = 11.83^o[/tex]  ...........4

from equation 3 and 4

[tex]T = \frac{86.6}{sin 11.83}[/tex]

T = 422.42 N

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