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3 years ago

7

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.

2 answers:

Aleks04 [339]3 years ago

5 0

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

$Tcos\theta = 413 N$ ........1

$176 = 413 sin 12.5 + Tsin\theta$ .......2

$Tsin\theta = 176 - 89.39 = 86.6$ .......3

divide equation 3 by 1

we get

$\theta = tan^{-1} (0.2096)$

$theta = 11.83^o$ ...........4

from equation 3 and 4

$T = \frac{86.6}{sin 11.83}$

T = 422.42 N

Whitepunk [10]3 years ago

4 0

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

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b) -2351.25 J

Explanation:

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$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

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$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

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