# A 200 lb man hangs from the middle of a stretched rope so that the angle between the rope and x axis is 5 degrees.Find the tension in the rope.

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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**