Question

Two ropes support a load of 539 kg. The two ropes are perpendicular to each other, and the tension in the first rope is 1.88 times that of the second rope. Find the tension in the second rope. The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.

Answer 1

Answer:

T₂ = 2482.34 N

Explanation:

Equations of balance of forces

Look at the force diagram in the attached graph:

∑Fx=0

T₂cos α  -T₁senα   = 0 Equation (1)

∑Fy=0

T₁cos α +T₂sinα-W =0 Equation(2)

Data

m=539 kg

g=  9.8 m/s²

W= m*g= 539 kg* 9.8 m/s²= 5282.2 N

T₁ = 1.88T₂

Problem development

in the equation (1)

T₂cos α-(1.88T₂)senα   = 0 We divided the equation by ( T₂cos α)

1 - (1.88)tanα   = 0

tanα= 1/(1.88)

tanα= 0.5319

$\alpha =tan^{-1} (0.5319)$

α = 28°

T₂=(1.88T₂)tanα

in the equation (2)

(1.88T₂)cos α+ T₂sinα - 5282.2 =0 We divided  the equation by cos α

1.88T₂+ T₂tanα - 5282.2/cos α =0

1.88T₂+ T₂tan(28°) - 5282.2/(cos 28°) =0

1.88T₂+ (0.53)T₂- 5982.46 =0

(2.41)T₂ = 5982.46

T₂ = 5982.46/(2.41)

T₂ = 2482.34 N