• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Determine the smallest diameter rope that can safely support

An artist friend of yours needs help hanging a sculpture from the ceiling. For artistic reasons, she wants 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 pounds per square inch of cross section. What size (diameter) rope should you buy?
Give your answer to the nearest 1/8 inch.

Fx= T2sin60 - T1sin(30)= 0
Fy= (T1cos30 + T2cos60) -mg = 0

T1= (T2sin60)/ sin30

mg= ((T2sin60)/ sin30) cos30 + T2cos60

mg = 1.5 T2 + .5T2
mg = 2T2
T2= mg/2 where T2 is the 60 degree
T1= T2sin60/sin30

If you change g=9.8 to 32
T2= 500 * 32/2= 8000

So pi(n/16)^2 * 4000 > or equal to T2
so pi(n/16)^2 * 4000 > or equal to 8000
pi(n/16)^2 > or equal to 2

What do I do now to find the needed diameter?????
PLEASE HELP!!!! I have to complete this by Tomorrow

Last edited:

Doc Al

Realize that you left out essential information when you presented the problem. For example, what's the strength rating of the rope?

Once you find the tension in each rope, you just need to find the minimum (I presume) diameter of rope that would survive such tension. If the rope is rated at X lbs/in^2, find the cross-sectional area of rope needed. Then convert that area to a diameter: [itex]A = \pi r^2 = \pi (d/2)^2[/itex].

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads