090928 Help

course Phy201

David,I have gotten completely out of sinc on some of these studies especially with the tension situations. I've read chpt 4 section 7 and I'm having a very hard time understanding the info. I guess I need some baby food info that will help me to understand. We have went from acceleration and velocity to the force of tension of which I'm not understanding. The problem is chpt 4 # 23 with the high walker on a rope. This is the problem;

Arlene is to walk across a high wire strung horizontally across two buildings 10.0 m apart. The sag in the rope when she is at the midpoint is 10.0 degrees. If her mass is 50.0 kg, what is the tension in the rope at this point?

I know F = m * a which is 50 kg * 9.8 m/s^2 = 490 N

the angle is 10 degrees and the midpoint is 5 m

so Fx = 490 cos 10 deg = 482.6 Newtons

Fy = 490 sin 10 deg = 85 Newtons

This is as far as I have been able to go. The book answer is 1410 Newtons of tension on the rope. I don't think my efforts are anywhere close to the correct direction to obtain the answer. So help!!!!

You are not far off, but there is no reason to assume that the tension in the rope is 490 N.

The tension acts to both sides, to the right end to the left. It is unknown so call it T.

The tension force to one side is directed at angle 10 degrees. The tension force to the other side is directed at angle 170 deg. So the total y force associated with the tension is

T sin(10 deg) + T sin(170 deg).

This is equal to the require 490 N so we have

T sin(10 deg) + T sin(170 deg) = 490 N.

You can solve this to obtain T.

Note also that sin(10 deg) and sin(170 deg) are equal. The tension to the right and the tension to the left have identical y components, as should be obvious from a sketch of the situation.