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StephenDoty
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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
Stephen
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
Stephen
Last edited: