- #1
ThePiGeek314
- 10
- 0
Homework Statement
Two similar questions here. I have the same question regarding solutions to both.
Calculate the tension in a horizontal strand of spider web if a spider of mass 8.00 × 10−5 kg sits motionless in the middle of it. The strand sags at an angle of 12° below the horizontal.
A monkey (m = 4 kg) is in a harness connected to a rope that goes up over a pulley on the ceiling. If the monkey pulls on the other end of the rope, it will go up. If it is climbing at a constant velocity, what is the tension in the rope?
Homework Equations
F = ma
W = ga
T = ∑Fn of all objects hanging from the rope
The Attempt at a Solution
In the first problem, the gravitational force on the spider is 7.8 * 10-4 N. Using right-triangle trig with the 12 degree given angle, the tension in the left half of the strand has to be 3.8 * 10-3 N. But that's only one half of the strand.
Tension in left half = tension in right half...because the spider is sitting right in the center
So the total tension should be 3.8 * 2 * 10-3 = 7.8 * 10-3 N.
Answer key says 1.89 * 10-3. 1.89 is a quarter of 7.8. Why is this? I know the spider is sitting in the middle of the web - I guess that means you might divide the 7.8 answer by 2, but not by 4.In the second problem, the gravitational force on the monkey is 39.2 N. The monkey is in equilibrium - moving with an acceleration of zero - so the tension on the rope must be 39.2 N.
But the correct answer is half that: 19.6 N. How can this be? Is it because the rope is part of a one-wheel pulley system, and that wheel divides the tension between the two rope segments? If it were a two-wheel system, would the tension then be 19.6 / 2 = 9.8 N?I think the answers to these questions are related. I can calculate the gravitational force, but I can't calculate tension with the force applied to the center of a rope, or in a pulley system. Can someone please explain this to me?