- #1
ssjdbz101
- 6
- 0
Homework Statement
A 25 kg object is being lifted by pulling on the ends of a 1.00 mm diameter nylon string that goes over two 3.10 m high poles that are 4.5 m apart, as shown in Figure 9-87. How high above the floor will the object be when the string breaks?
*see attached picture*
The foreces on the rope (called Fa and Fb) are equal.
The Attempt at a Solution
I know that the forces to fracture the rope are in the directions of the rope. In this case, Fa and Fb, and they have to be greater then the tensile strength of the nylon, which is 1570N.
So, the Forces in the y direction are the Fg of the box (25kg x 9.8m/s^2)=245N, and the combined forces of the the string (Fay and Fby) = 245N. If the forces Fa and Fb are equal, then the forces Fay and Fby are equal, each at 122.5N, correct?
I don't really know how to set-up the problem.
If Fa+Fb>1570N, then I see Fa and Fb=786N (added up to be greater than 1570N).
Now, using the Pythagorean Thm, I know the hypotenuse and a leg of the right triangles made by the forces on the rope. I find the angle to be 8.96 degrees.
I then look at the right triangle made by the lengths of the poles. If the bottom leg is 1/2 the total length, 4.5m, then a leg on on triangle is 2.25m. To find the heght at 8.96 degrees, I take the tan(8.96) x 2.25m = 0.35m. But I know that is not the right answer.
What am I doing wrong? (Sorry for having a lot to read, but this is how I did the problem)
Thanks.