Question

A uniform drawbridge must be held at a 37 ∘ angle above the horizontal to allow ships to pass underneath. the drawbridge weighs 45,000 n and is 14.0 m long. a cable is connected 3.5 m from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place. part a what is the tension in the cable?

Answer 1

Answer:

T = 119638 N

Explanation:

Since the drawbridge is at equilibrium and not moving at the given situation

so we can apply torque balance in this case

so here torque due to weight is counterbalanced by the torque due to tension in the string

Torque due to weight is given as

$\tau_1 = mg(\frac{L}{2}cos37)$

torque due to tension in string

$\tau_2 = T(3.5 sin37)$

now by torque balance equation we have

$T(3.5sin37) = mg(\frac{L}{2}cos37)$

$T(2.1) = 45000(7)(0.8)$

$T = 119638 N$

Answer 2

Tension may be defined as the pulling power transferred axially through a cable, string, chain, or alike one-dimensional unceasing object, or by separately end of a rod.

To compute for tension:

Sum the moments about the pivot: 

ΣM = 0 = T * 3.5m * sin37 º - 45000N * 7.0m * cos37º 
tension T = 119 434 N