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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.

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Answer:

Tā‚‚ = 2482.34 N

Explanation:

Equations of balance of forces

Look at the force diagram in the attached graph:

āˆ‘Fx=0

Tā‚‚cos Ī± Ā -Tā‚senĪ± Ā  = 0 Equation (1)

āˆ‘Fy=0

Tā‚cos Ī± +Tā‚‚sinĪ±-W =0 Equation(2)

Data

m=539 kg

g= Ā 9.8 m/sĀ²

W= m*g= 539 kg* 9.8 m/sĀ²= 5282.2 N

Tā‚ = 1.88Tā‚‚

Problem development

in the equation (1)

Tā‚‚cos Ī±-(1.88Tā‚‚)senĪ± Ā  = 0 We divided the equation by ( Tā‚‚cos Ī±)

1 - (1.88)tanĪ± Ā  = 0

tanĪ±= 1/(1.88)

tanĪ±= 0.5319

[tex]\alpha =tan^{-1} (0.5319)[/tex]

Ī± = 28Ā°

Tā‚‚=(1.88Tā‚‚)tanĪ±

in the equation (2)

(1.88Tā‚‚)cos Ī±+ Tā‚‚sinĪ± - 5282.2 =0 We divided Ā the equation by cos Ī±

1.88Tā‚‚+ Tā‚‚tanĪ± - 5282.2/cos Ī± =0

1.88Tā‚‚+ Tā‚‚tan(28Ā°) - 5282.2/(cos 28Ā°) =0

1.88Tā‚‚+ (0.53)Tā‚‚- 5982.46 =0

(2.41)Tā‚‚ = 5982.46

Tā‚‚ = 5982.46/(2.41)

Tā‚‚ = 2482.34 N

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