a 120 kg uniform beam is attached to a vertical wall at one end is supported by a vertical cable at the other end. calculate the magnitude of the tension in the wire if the angle between the beam and the horizontal is delta=27 degrees

t₁ = 588 [N]

Explanation:

There are 2 forces acting in the free body diagram.

The weight of the beam   w = 120*9,8  

w = 1176 [N]

Force w= 1176 [N] ( the weight of the beam acting in the center of mass) and the tension t₁ in the vertical free end. As the system is in equilibrium, the sum of torque related to the attached point to the vertical wall must be equal to 0 then. Both forces are vertical forces therefore

∑T = f₁ * L/2 - t₁ * L = 0

1176*L/2 - t₁ * L = 0

1176/2 -t₁ = 0

t₁ = 588 [N]

wavelength divides in half and speed stays the same.