1B. The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights. a) What is the tension TA in cord A? b) What is the tension TB in cord B? c) What is the tension TC in cord C? 65 A C 1000 dyn 125° 500 dyn B 6 65° 120 500N fore the wo! ** B .. 95⁰ loooN In the Horisontal Dix. Net Foxice =0 So, esin 15⁰- B sin 80 = 0 on, c sin 15° = B sin 80° OX c= 3.8 x8 In the Vertical Dik. Net fasa = 0 50, e cos 15° + 8 eos 80 = 1000 on, (3.8x B) x cos 15° OP, 010, B= 260. 14 Newton c = 3.8x8 = In the Horizontal Dise. Net fosce =0 50, A sin 25⁰ = 8 sin 80 on, A = 2.33xB = 606.12 Newton. 9 for the knot J: 75° с Bx [3.67+ 0.174] = 1000 TA= 606-12 N + B cos 80 = 1000 100° 15 '80° By simple geometrical Calculations. Bcas 80 Ć cos Isº 4 8 15° →e sin 15° B sin 80° 988.55 Newton. 80 Asin 20 To = 260.14N, Te = 988.55N Am. 1000 N Free Body Diagram AB 100⁰ > Bsingo 500N B Pree Body Diagram
1B. The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights. a) What is the tension TA in cord A? b) What is the tension TB in cord B? c) What is the tension TC in cord C? 65 A C 1000 dyn 125° 500 dyn B 6 65° 120 500N fore the wo! ** B .. 95⁰ loooN In the Horisontal Dix. Net Foxice =0 So, esin 15⁰- B sin 80 = 0 on, c sin 15° = B sin 80° OX c= 3.8 x8 In the Vertical Dik. Net fasa = 0 50, e cos 15° + 8 eos 80 = 1000 on, (3.8x B) x cos 15° OP, 010, B= 260. 14 Newton c = 3.8x8 = In the Horizontal Dise. Net fosce =0 50, A sin 25⁰ = 8 sin 80 on, A = 2.33xB = 606.12 Newton. 9 for the knot J: 75° с Bx [3.67+ 0.174] = 1000 TA= 606-12 N + B cos 80 = 1000 100° 15 '80° By simple geometrical Calculations. Bcas 80 Ć cos Isº 4 8 15° →e sin 15° B sin 80° 988.55 Newton. 80 Asin 20 To = 260.14N, Te = 988.55N Am. 1000 N Free Body Diagram AB 100⁰ > Bsingo 500N B Pree Body Diagram
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Can you explain to me how did the expert came up with their conclusions on each angle. Specially the 75 on the upper part. I am confused how did they arrived with those values when 125 and 65 degrees are only given in the problem.
1st pic is the problem and next is the solution annd the FBD. Please, I need ur help immediately. Thank you so much!
Transcribed Image Text:1B. The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights.
a) What is the tension TA in cord A?
b) What is the tension TB in cord B?
c) What is the tension TC in cord C?
65
A
C
1000 dyn
125°
500 dyn
B
![6
65°
120
500N
fore the wo! **
B
..
95⁰
loooN
In the Horisontal Dix. Net Foxice =0
So,
esin 15⁰-
B sin 80 = 0
on,
c sin 15°
= B sin 80°
OX
c= 3.8 x8
In the Vertical Dik. Net fasa = 0
50,
e cos 15° + 8
eos 80 = 1000
on, (3.8x B) x cos 15°
OP,
010, B= 260. 14 Newton
c = 3.8x8 =
In the Horizontal Dise. Net fosce =0
50,
A sin 25⁰ =
8 sin 80
on,
A =
2.33xB
= 606.12 Newton.
9
for the knot J:
75°
с
Bx [3.67+ 0.174] = 1000
TA= 606-12 N
+ B cos 80 = 1000
100°
15
'80°
By simple geometrical Calculations.
Bcas 80
Ć cos Isº
4
8
15°
→e sin 15°
B sin 80°
988.55 Newton.
80
Asin 20
To = 260.14N, Te = 988.55N
Am.
1000 N
Free Body Diagram
AB
100⁰
> Bsingo
500N
B
Pree Body Diagram](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2989cfcb-62a0-4dd3-961e-3ec0c538fc35%2F2e1dd8c0-af07-489c-9764-e0ca5a52ca98%2Fb6um2n_processed.png&w=3840&q=75)
Transcribed Image Text:6
65°
120
500N
fore the wo! **
B
..
95⁰
loooN
In the Horisontal Dix. Net Foxice =0
So,
esin 15⁰-
B sin 80 = 0
on,
c sin 15°
= B sin 80°
OX
c= 3.8 x8
In the Vertical Dik. Net fasa = 0
50,
e cos 15° + 8
eos 80 = 1000
on, (3.8x B) x cos 15°
OP,
010, B= 260. 14 Newton
c = 3.8x8 =
In the Horizontal Dise. Net fosce =0
50,
A sin 25⁰ =
8 sin 80
on,
A =
2.33xB
= 606.12 Newton.
9
for the knot J:
75°
с
Bx [3.67+ 0.174] = 1000
TA= 606-12 N
+ B cos 80 = 1000
100°
15
'80°
By simple geometrical Calculations.
Bcas 80
Ć cos Isº
4
8
15°
→e sin 15°
B sin 80°
988.55 Newton.
80
Asin 20
To = 260.14N, Te = 988.55N
Am.
1000 N
Free Body Diagram
AB
100⁰
> Bsingo
500N
B
Pree Body Diagram
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