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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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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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