Respuesta :
Answer: T1 = 6554.6N
Explanation:
Given;
Weight w = 5500N
Cable 1 = T1 acting on the horizontal
Cable 2 = T2 acting 40° to the horizontal
To solve this kind of question we must take the horizontal and vertical component of force at equilibrium.
Taking the summation of forces on the horizontal (x axis)
T1 - T2(cos40°) = 0 .....1
Summation of forces on the y axis (vertical)
T2(sin40°) - W = 0 .....2
From equation 2
T2(sin40) = W
T2 = W/sin40 .....3
From equation 1
T1 = T2(cos40)
Substituting T2 = W/sin40
T1 = (W/sin40)(cos40)
T1 = 5500N(cos40/sin40)
T1 = 6554.6N
The horizontal and vertical component of forces, are at equilibrium. The tension acting on the horizontal in cable one is 6554.6 N.
W = Weight = 5500 N
T1 - Tension acting on the horizontal in cable one.
T2 - Tension acting [tex]\bold{40^o}[/tex] to the horizontal surface in cable 2
Take the horizontal and vertical component of force at equilibrium.
Sum up forces on the X-axis
T1 - T2 (cos40°) = 0 .........1
Summation of forces on the Y-axis
T2 (sin40°) - W = 0 ..........2
From equation 2
T2 (sin40) = W
[tex]\bold {T2 = \dfrac W{sin40} }[/tex] ......3
From equation 1
T1 = T2(cos40)
Put the value of T2 in equation 1,
[tex]\bold {T1 = \dfrac W{(sin40)} (cos40)}}\\\\\bold {T1 = \dfrac {5500}{(sin40)} (cos40)}}\\\\}\bold {T1 = 6554.6\ N}[/tex]
Therefore, the tension acting on the horizontal in cable one is 6554.6 N.
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