When a cable is passed over a pulley on the pier and is stayed at the back, the pier will be subjected to net horizontal force which is given by (where H is horizontal tension in the cable, α is angle made by the cable with vertical and β is angle made by the backstay with vertical at pier.)
- AH (1 − sinα · cosecβ)
- BH (1 − sinβ · cosecα)Correct
- CH (1 + sinα · cosecβ)
- DH (1 + sinβ · cosecα)
Solution
Concept
Net horizontal force on a pier from a cable and its backstay
Formula
F = H − T sinβ, T = H / sinα ⇒ F = H (1 − sinβ cosecα)
Step-by-step solution
1. The cable tension T = H / sinα (the horizontal component is T sinα = H, with α measured from the vertical). 2. Over a smooth pulley the backstay tension equals T. 3. Horizontal component of the backstay pull = T sinβ = H sinβ cosecα, acting in the direction opposite to H. 4. Net horizontal force = H − H sinβ cosecα = H (1 − sinβ cosecα).
Explanation
Both the cable and the backstay pull on the top of the pier, in opposite horizontal directions. The net horizontal force on the pier is the difference of their horizontal components.
Why the other options are wrong
Plus signs would add the two horizontal pulls although they act on opposite sides. Options with sinα cosecβ interchange the angles.
Shortcut
T = H cosecα; backstay horizontal = T sinβ; net = difference.
Common mistake
Adding the two horizontal components.
Correct answer: B) H (1 − sinβ · cosecα)