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ryzh [129]
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.

Physics
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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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

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