Each cord can sustain a maximum tension of 510 N Determine t

Each cord can sustain a maximum tension of 510 N. Determine the largest mass of pipe that can be supported. Express your answer to three significant figures and include the appropriate units.

Solution

Solution :

Assume that P is the weight of pipe .
AP is vector of this weight
AK, AI=BM ,BJ,BL are the forces apply to these cords AE,AB,BC,BD respectly .

The angle PAI = 90º - 60º = 30º ( because KAP = 90* and BAI is straight angle)
In the right triangle :

AI= BM = AP/sin60º x AK = PI= AP/tan60º

And then we have:

<BLM = <BJM = arc sin 3/5 = 36.87º

( < means angle )

<MBJ = 60º---> BMJ = 180º - ( 36.87º + 60º) = 83.13º

Apply sine method :

BJ/sin 83.13º = JM/sin60º = BM/sin 36.87º

----> BJ = 1.65 BM ; BL = 1.44 BM

We prove above , BM= AP/sin60º

Therefore the force applies to BC cord is biggest ,and it equals 1.65BM =1.65 X AP/sin 60º = 1.9AP

Because Each cord can sustain a maximum tension of 500 N ,so the maximum weight of pipe is :

1.9 AP = 510 N ------> AP = 268.42N

With the gravity is 9.81 ,the maximum mass of pipe is :

m = 268.42/9.81 =27.36 kg