Question

A rope, assumed massless, is stretched horizontally between two supports that are 3.23 m apart. When an object of weight 3470 N is hung at the center of the rope, the rope is observed to sag by 32.5 cm. What is the tension in the rope?

Answer 1

Answer:

T= 8793 .04 N

Explanation:

We calculate the angle (β) formed by the rope with the horizontal after hanging the weight:

tan (β ) = V / H  Formula (1)

Where :

tan (β) : β angle tangent

H : horizontal distance from a support to the center

V : vertical displacement of the rope when the weight is hung

Data

H = 3.23 m / 2 = 1.615 m =  161.5 cm

V = 32.5 cm

β angle calculation

We replace data in the formula (1)

tan (β) = 32.5 cm /161.5 = 0.20124

$\beta = tan^{-1} (0.20124)$

β = 11.38°

Forces acting in the center of the rope

W: Weight of the object : In vertical direction  (y) and downwards

T₁ : Rope tension to the left and upwards

T₂ : Rope tension to the right and upwards

x-y components of the T₁ and T₂

T₁ = T₂ = T because it is the same rope

T₁x = T₂x = T*cos( 11.38°)

T₁y = T₂y = T*sin( 11.38°)

Equilibrium equation of forces in vertical direction inthe center of the rope

∑Fy = 0

2T₁y - W = 0

2 ( T*sin( 11.38°) - 3470 N  = 0

2T*sin( 11.38°)  = 3470 N

0.3946 * T = 3470 N

$T= \frac{3470 N}{0.3946}$

T= 8793 .04 N