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When the man is hanging, the 2 sections of the rope are symmetrical, with respect to the man.
The tensions, T1 and T2, from the 2 sections of the rope, have the same magnitude.
Since the system is in equilibrium the sum of all forces in the horizontal direction is zero.
T1*cos 5 - T2*cos5 = 0
But T1 = T2 = T
Also, the sum of all forces in the vertical direction is zero.
T1*sin 5 + T2*sin 5 - 200 = 0
We'll add 200 both sides:
sin 5*(T1+T2) = 200
2T*sin 5 = 200
We'll divide by 2:
T*sin 5 = 100
T = 100/sin 5
T = 100/0.0087
T = 1150 lbs
We notice that the tension in the rope is over five times the weight of the man.
The man hangs from the middle of the rope so that the angle between the rope and the x axis is 5 degree.
The force acting downwards due to the man of weight 200 lb is 200*4.45 N = 890 N
The tension (T) in the rope can be divided into two equal parts with each pulling at the center equally. Each of these can be divided into two components: a vertical component of T sin 5 and a horizontal component of T cos 5.
The vertical components point upwards and their sum is equal to the force of 890 N acting downwards. Therefore we have that 2*T sin 5 = 890.
=>T sin 5 = 890/2
=>T sin 5 = 445
=> T = 445 / sin 5
=> T = 5105 N
=> T = 5.105 KN
or 1147 lb
Therefore the tension in the rope is 1147 lb
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