A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes, fastened at different heights, make angles of 50Â° and 38Â° with the horizontal. find the tension in each rope and the magnitude of each tension. (let t2 and t3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. round all numerical values to two decimal places.)

Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

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abidemiokin abidemiokin · Answer 2 · 2021-11-02T16:05:09+01:00

The measure of the tension T2 and T3 to two decimal places is 271.22N and 332.52N respectively.

To get the tension [tex]T_2[/tex] and [tex]T_3[/tex], we will need to resolve the force into the horizontal and vertical components as shown:

For the horizontal component,

[tex]T_2cos 50 = T_3cos38\\0.6428T_2=0.7880T_3\\T_2=\frac{0.7880T_3}{0.6428}\\T_2= 1.226T_3[/tex]

For the vertical component:

[tex]T_2sin50 + T_3sin38=430\\1.266T_3sin50+0.6156T_3=430\\0.9698T_3+0.6156T_3=430\\1.5854T_3=430\\T_3=\frac{430}{1.5854}\\T_3 =271.22N[/tex]

Since

[tex]T_2= 1.226T_3\\T_2= 1.226(271.22)\\T_2=332.52N[/tex]

Hence the measure of the tension T2 and T3 to two decimal places is 271.22N and 332.52N respectively.

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