What is the tension of the rope at this point??

Question

Arlene is to walk across a high wire strung horizontally between two buildings 13.0 m apart. The sag in the rope when she is at the midpoint is 10.0°. If her mass is 58.0 kg, what is the tension in the rope at this point?.....N

Answer 1

consider the tension on the right side of arlene T1 and the left side T2 (you will eventually see these are equal but just to explain how to work the problem we will consider them separately, they are only equal because Arlene is directly in the center)

Sum of the forces in the X direction = 0 and are:
T1cos(10)-T2cos(10)=0

Sum of the forces in the Y direction = 0 and are:
T1sin(10)+T2sin(10)-58kg(9.8 m/s2)=0

Solve for T1 in the first equation you will find that T1=T2 which I will now simply call T
Therefore the second equation becomes
2Tsin(10)-58(9.8)=0

T = 1638.64

If you always break these problems down into forces in the X and Y directions setting each equal to zero they become a matter of just plugging in numbers. If a third dimension is added you simply do the same process but also break them down in the Z direction as well (which also will always equal 0)

Answer 2

You should start by drawing a force diagram.
3 Forces
Weight down
Tension (10 degrees up from left)
Tension (10 degrees up from right)

By symmetry, the horizontal forces on Arlene's toe cancel

Vertical force on Arlene's toe
= 2T sin theta - mg = 0

You have her weight down and each side of the rope pulling up with a vertical component of the tension.

So T = mg / (2 sin theta)

hey give you m and theta. You know g. The distance between buildings doesn't matter. Plugnchug.

Answer 3

Many problems of this sort can be attacked by listing all the forces that are acting on an object (Arlene in this case), and making equations for the forces.

There are three forces acting on Arlene:

Gravity (force = mg)
The tension on the rope pulling on one side (T)
The tension on the rope pulling on the other side (T)

Tension always pulls in the direction of the rope. So you can figure out what part of the tension pulls upward, and what part pulls sideways.

Upward pull (due to each half of rope) = (T)sin10°
Sideways pull (due to each half of rope) = (T)cos10°

Now, Arlene is not accelerating, which means all the forces on her cancel out. That means:

upward forces = downward forces

or

2(T)sin10° = mg

(I multiplied the left half by "2" because there are 2 pieces of rope pulling up.)

From this, you can solve for T.