A bale of weight 24 kN hangs from a steel cable, (a) Determine the tension T in the cable if the bale is held to one side with a horizontal force of 1.2 kN. (b) If the cable is 13 m long, what horizontal force is required to hold the bale a horizontal distance of 5 m from the vertical line through the point of support?1kN=1000 N
To determine the tension in the cable, we need to use the concept of equilibrium. In this case, the sum of the forces acting in the vertical direction should equal zero.
(a) Given:
Weight of the bale (W) = 24 kN
Horizontal force (F) = 1.2 kN
To find the tension (T) in the cable, we can set up the equation:
T - W - F = 0
Substituting the values, we have:
T - 24 kN - 1.2 kN = 0
T - 25.2 kN = 0
Rearranging the equation, we find:
T = 25.2 kN
Therefore, the tension in the cable is 25.2 kN.
(b) Given:
Length of the cable (L) = 13 m
Horizontal distance (d) = 5 m
To find the horizontal force required, we can use the concept of similar triangles.
Let's set up the equation using the ratios of corresponding sides of similar triangles:
Horizontal force / Vertical distance = Length of cable / Horizontal distance
F / W = L / d
Substituting the values, we have:
F / 24 kN = 13 m / 5 m
F / 24 kN = 2.6
Rearranging the equation, we find:
F = 2.6 * 24 kN
F = 62.4 kN
However, you mentioned that 1 kN is equal to 1000 N, so let's convert the result to Newtons:
F = 62.4 kN * 1000 N/1 kN
F = 62,400 N
Therefore, the horizontal force required to hold the bale 5 m from the vertical line through the point of support is 62,400 N.