Dynamics of a Cord: Solving Tension for Arlene

[ciubba]

Homework Statement

Arlene is to walk across a “high wire” strung horizontally between two buildings 10.0 m apart. The sag in the rope when she is at the midpoint is 10.0°, as shown in Fig. 4–47. If her mass is 50.0 kg, what is the tension in the rope at this point?

Homework Equations

The definitions of sine and cosine.
$$\sum F=ma$$

The Attempt at a Solution

I'm really confused by how the tension on the cord could be 1410N, when the rope-walker only exerts a 'weight' of 490N. Evidently, there must be some aspect of cord-tension that I do not understand.

 

[BvU]

Perhaps there are angles involved... tensions in wires and such can only be in the direction of the wire. Gravity, however, is pulling downwards. Experiment with a weight hnging from a rope you hold with your amrs spread wide.

Did you learn about decomposing a force in given direction into components in other directions ?

 

[SteamKing]

If you draw a simple sketch of the high-wire artist and the sagged line, and then insert the forces, the solution should become apparent. Remember, at the point where the artist stands on the wire, the sum of the forces in the horizontal and vertical directions must equal zero, since everything is in equilibrium.

 

[ciubba]

If you draw a simple sketch of the high-wire artist and the sagged line, and then insert the forces, the solution should become apparent. Remember, at the point where the artist stands on the wire, the sum of the forces in the horizontal and vertical directions must equal zero, since everything is in equilibrium.

Ah, thank you. 490/2=245N

245/sin(10)=1410N.

 

[HalfLostAndSearching]

I would approach this problem by first analyzing the forces acting on the cord. The two main forces are the weight of Arlene, which is acting downwards, and the tension in the rope, which is acting upwards. The fact that the cord is sagging at a 10.0° angle indicates that the tension is not equal to the weight, but is instead greater.

To solve for the tension, we can use the equation \sum F=ma, where \sum F is the net force acting on the cord, m is Arlene's mass, and a is the acceleration of the cord (which we will assume to be zero since the cord is not moving). We can also use the definitions of sine and cosine to relate the angles in the problem to the forces.

First, let's consider the vertical forces. The weight of Arlene, 490N, is acting downwards. The tension in the cord is acting upwards at an angle of 10.0° from the vertical. Using the definition of cosine, we can determine that the vertical component of the tension is Tcos(10.0°).

Now, let's consider the horizontal forces. The tension in the cord is acting to the right, and is offset by the weight of Arlene acting to the left. Using the definition of sine, we can determine that the horizontal component of the tension is Tsin(10.0°).

Since the cord is not moving, the net force in the vertical direction must be zero. This means that Tcos(10.0°) = 490N. Solving for T, we get T = 490N / cos(10.0°) = 502N.

Therefore, the tension in the rope at the midpoint, where Arlene is walking, is 502N. This is greater than her weight of 490N, which is necessary to keep the cord in equilibrium and prevent her from falling.