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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

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