The weight of the horizontal wall section is W D 20,000 N. Determine the tensions in the cables AB, AC, and AD.

Question

The weight of the horizontal wall section is W = 20,000 N. Determine the tensions in the cables AB, AC, and AD. 

Answer 1

Set the coordinate origin at A with axes as shown. The upward force, T, at point A will be equal to the weight, W, since the cable at A supports the entire wall. The upward force at A is T = W k. From the figure, the coordinates of the points in metre are

A(4,6,10) B(0,0,0) C(12,0,0) and D(4,14,0).

The three unit vectors are of the form

where I takes on the values B, C, and D. The denominators of the unit vectors are the distances AB, AC, and AD, respectively. Substitution of the coordinates of the points yields the following unit vectors: 

The equilibrium equation for the knot at point A is

T + T_AB + T_AC + T_AD = 0.

From the vector equilibrium equation, write the scalar equilibrium equations in the x, y, and z directions. We get three linear equations in three unknowns. Solving these equations simultaneously, we get

T_AB = 9393 N, T_AC = 5387 N, and T_AD = 10,977 N.