Question

A uniform beam of length x = 1.0 m and mass 10 kg is attached to a wall and supported by a cable. The wall is vertical. The beam is horizontal. The cable attaches at the end of the beam farthest from the wall at angle of 300 to the horizontal. The other end of the cable attaches to the wall above the point where the beam is attached to the wall. (Thus, the beam, wall, and cable form a right triangle with the cable as the hypotenuse.) The beam is free to pivot at the point where it attaches to the wall. What is the tension in the cable ?

Answer 1

Answer:

T = 56.6 N

Explanation:

To solve this exercise we will use Newton's second law, for translational and rotational equilibrium

X axis

    Σ F = 0

   Fₓ -Tₓ = 0

  Tₓ = Fₓ

Y Axis

  $F_{y}$ - W + $T_{y}$= 0

Rotational

   Σ τ = 0

Spinning point on the wall where the bar rests, the counterclockwise tour is positive

    -Tₓ L + W L / 2 = 0

     Tₓ = W / 2

let's look for the stress components with trigonometry

    sin 30 = Ty / T

    cos 30 = Tₓ / T

    $T_{y}$ = T sin30

    Tₓ = T cos 30

We calculate

    T cos 30 = W / 2

    T = mg / (2 cos 30)

    T = 10 9.8 / (2 cos 30)

    T = 56.6 N

Answer 2

The tension in the cable is mathematically given as

T = 56.6 N

The tension in the cable

Generally the equation for the Newton's second law  is mathematically given as

F=ma

Therefore

For X axis  

Tx = Fx

For Y Axis

Fy - W + Ty= 0

Hence,    

Tx = W / 2

Resolving the stress components we  have

T_{y} = T sin30

Tₓ = T cos 30

solving

T cos 30 = W / 2    

T = mg / (2 cos 30)    

T = 10 9.8 / (2 cos 30)    

T = 56.6 N

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