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A block of mass M = 15.0 kg is suspended at rest by two strings attached to walls, as shown in the figure. The left string is horizontal and the right string makes an angle _ = 50.0Á with the horizontal. What is the tension in the left string? (Assume the ring where the strings come together is massless.)

Respuesta :

Edufirst
Forces that appear in the free body diagram

1) T2: horizontal tension to the left

2) T1: tension to the right, making angle 50° with the horizontal

     Vertiical component of T1: T1y = T1*sin(50°)
     Horizontal component of T1 = T1x = T1. cos(50°)

3) Weight of the mass: 50 kg * 9.8 m/s^2 = 490 N

4) Equilibrium analysis

Net vertical forces = 0 => T1y = weight => T1*sin(50°) = 490 N => T1 = 490 N / sin (50°) = 639.7 N

T1 = 639.7 N

Net horizontal forces = 0 => T2 = T1x = T1*cos (50°) = 639.7 N * cos(50°) = 411.2 N

Answer: 411.2 N

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