Physics! plese help

A 1320-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the figure, find (a) the magnitude of the tension in the wire and the magnitudes of the (b) horizontal and (c) vertical components of the force that the wall exerts on the left end of the beam.

From the wall to the beam creates a 50 degree angle. From the beam to the crate is a 30 degree angle.

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  1. The drawing shows the beam and the five forces that act on it: the horizontal and vertical components S(x) and S(y) that the wall exerts on the left end of the beam, the weight W(b) of the beam, the force due to the weight Wc of the crate,
    and the tension T in the cable. The beam is uniform, so its center of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the torques about any axis of rotation must be
    zero ( Σ τ = 0) , and the sum of the forces in the horizontal and vertical directions
    must be zero ( Σ F(x)= 0, Σ F(y)= 0) . These three conditions will allow us to determine the magnitudes of S(x), S(y), and T.
    We will begin by taking the axis of rotation to be at the left end of the beam. Then
    the torques produced by S(x) and S(y) are zero, since their lever arms are zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, T, in it. Setting the sum of the torques produced by the three forces equal to zero gives (with L equal to the length of the beam)
    Σ τ = − W(b){0.5• L•cos30°} − W(c) •{Lcos30°} + T• {L•sin80°} = 0.
    Algebraically eliminating L from this equation and solving for T gives
    T = [W(b){0.5• L•cos30°} + W(c) •{Lcos30°}]/sin80º =
    ={1320• 0.5 •cos30°+1960• cos30°}/sin 80 º=2883 N.

    Since the beam is in equilibrium, the sum of the forces in the vertical direction
    must be zero:

    Σ F(y) = + S(y) − W(b) − W(c) + T•sin50° =0
    Solving for S(y) gives
    S(y) =W(b)+W(c)-T•sin50 =1320+1960-2883•sin50°=1071 N.
    The sum of the forces in the horizontal direction must also be zero:
    Σ F(x)= + S(x) − T• cos50° = 0.
    so that
    S(x) = T• cos50° =2883•cos50° =1853 N.

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