Horizontal pull in a parabolic cable (light suspension bridge) A light suspension bridge has span l and central dip y. Each cable carries a total uniformly distributed load w (total over the span on that cable). What is the horizontal pull H at each support?

Introduction / Context:The horizontal component of cable tension in a parabolic cable under uniform load controls anchor design and sag selection. The classic relationship links span, sag (dip), and load.

Given Data / Assumptions:

• Each cable carries a total load w over the span.
• Parabolic cable approximation under uniform horizontal projection of load.
• Span = l, central dip (sag) = y.
• Statics of a flexible, inextensible cable.

Concept / Approach:For a parabolic cable with uniform load per unit horizontal length w_h, the horizontal pull is H = w_h * l^2 / (8 y). If the problem states w as the total load on a cable over the span, then w_h = w / l, giving H = (w / l) * l^2 / (8 y) = (w l) / (8 y).

Step-by-Step Solution:

Verification / Alternative check:Units: w is force, l/y is dimensionless, so H has units of force—consistent. If w were given per unit length instead, H would be w l^2 / (8 y), which corresponds to a different interpretation.

Why Other Options Are Wrong:

• (w l^2)/(8 y): Treats w as load per unit length, not as total load.
• (w y)/(8 l) and (8 w y)/l^2: Wrong dependence on y (predicts larger H for larger sag, opposite of reality).
• w l: Ignores sag entirely.

Common Pitfalls:Confusing w as total versus per-unit-length; remember to convert to w_h when applying the parabolic formula.

Final Answer:H = (w l) / (8 y)