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.