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.

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.