Originally Posted by **toranc3** An artist friend of yours needs help hanging a 500 pounds sculpture from the ceiling. For artistic reasons, shewants to use just two ropes. One will be from vertical, the other . She needs you to determine the smallest diameter rope that can safely support this expensive piece of art. On a visit to the hardware store you find that rope is sold in increments of diameter and that the safety rating is 4000 pounds per square inch of cross section.
What size (diameter) rope should you buy?Give your answer to the nearest 1/8 inch
I did my summations of forces. T1 is 30 degrees from vertical and T2 is 60 degrees from vertical.
Fy=may
T1y + T2y - W =0
Fx=max
T2x - T1x=0
T1= 1539N
T2= 889N.
I am stuck for the next part.
T1 |

Sometimes we have to use crazy units (pounds, psi, inch). The one to watch out for is "pound," which is almost always force (weight), not mass. Rather that switch back and forth, I would keep the question in pounds, or you could say "units of 4.4482 N." So when I solve for the two tensions I get

T1 = 433# (= 1926 N)

T2 = 250# (= 1112 N)

Again, you haven't shown intermediate steps so I don't know where we parted ways. In any case, it is T1 we have to worry about.

The tensile strength has units of #/in.^2. dividing T1 by that number will give the required cross section of the rope.

A = (433#)/(4000 #/in.^2) = 0.108 in.^2

Find the

*diameter*, and round UP to the next larger multiple of 0.125 in.

Then, because I am a physicist and "believe" the safety factors given in the specifications, the rope will break, marking the end of a friendship. Meanwhile, the engineer who always puts in an additional factor of 3 for safety, will succeed and win the heart of the beautiful sculptress.