First Condition of Equilibrium

by acsin92
Tags: condition, equilibrium
 P: 4 http://img369.imageshack.us/my.php?i...eng0038rn2.jpg Question: To move a heavy crate across a floor, one end of a rope is tied to it and the other end is tied to a wall 30 ft. away. When a force of 100 lbs. is applied to a midpoint of the rope, the rope stretches so the midpoint moves to the side by 2ft. What is the force on the crate? I tried using -T+w cos30+w cos 30 to get the force on the crate but now I know its wrong. Then looking on the diagram, I cant seem to find what angle to use, since I think its one important detail on how to solve the problem. All I know is that the 100 lbs. is the tension applied to the rope and the weight of the crate was not given. Please help.
Mentor
P: 40,264
 Quote by acsin92 All I know is that the 100 lbs. is the tension applied to the rope and the weight of the crate was not given.
100 lbs is the force applied to the rope, not the tension in the rope. (The tension is what you're trying to find.) You don't need the weight of the crate.

Analyze a section of rope at the very midpoint. What forces act on it? Hint: Both sides of the rope pull at that point.

To find the angle, look at the right triangle formed between crate, original midpoint, and new midpoint. What's the horizontal distance from crate to midpoint? That's one side. What's the vertical deflection from horizontal? That's another side.
 P: 4 yeah i got it. T = 31.61N.

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