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Please check if I completed this question correctly. If I solved it correctly. And the answer is correct   Image 1: The question Image 2: My work

Please check if I completed this question correctly. If I solved it correctly. And the answer is correct   Image 1: The question Image 2: My work

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Please check if I completed this question correctly. If I solved it correctly. And the answer is correct

 

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Hit A wire # 2 - 0₁ = 40° 0₂ = 50° a) Ti Cos 0₁ T₂ cos ₂ = 0 27 T₁ Cos 6₁ = T₂ Cos 0₂ Ti Costi ( T2 -> T₂ = COS 6₂ 8 kat m Hence, Ti 8. <wire #1 -▷ Decoration Diagram and Ti Sin 0₁ + T₂ Sinf₂ = • mg Sine Ti sinfi + + Cos ₁ we are going to use equation later. Ti Costi Sine₂ = Cos 0₂ Jir: 62 Cos 02 mg Hence, T₁ = sin Q₁ + Cos 0₁, tanAz = mg 8x9.8 Hence Ti = Sin 40° + cos 40⁰ tan 50° 77 SO.39455 N Cos 40° X 12 = 50.39455 x COS 450 cos so° -T₂ = 60.05788 N AL 162 a) Now given 0₁ = 40⁰ and £₂ = 50° and m= Free Body Diagram 314 01 mg To cos d₂ [Bj T2 T₂ R 0₂ 0101 = 8kg Tr my b) From A ABC sing₂ = I yo Apply Newtons second law T₂ cos 6₂ = man Ti Cost1 - Tz costz Tisins + T₂ SinG - nug = may Now l=0.3m T₁ = Sinfi + T2 sin 02 Ii Cose 1 DC = Sin 40° BC =) 0-3 = sin 40° => DC = 0.19284 1 BC=0.3m The tension for wire I is T₁ = 50.39.N and wire #2 its tension is T2 = 60.06 N b): The decoration is suspended 0.19 meter below the ceiling
Transcribed Image Text:Hit A wire # 2 - 0₁ = 40° 0₂ = 50° a) Ti Cos 0₁ T₂ cos ₂ = 0 27 T₁ Cos 6₁ = T₂ Cos 0₂ Ti Costi ( T2 -> T₂ = COS 6₂ 8 kat m Hence, Ti 8. <wire #1 -▷ Decoration Diagram and Ti Sin 0₁ + T₂ Sinf₂ = • mg Sine Ti sinfi + + Cos ₁ we are going to use equation later. Ti Costi Sine₂ = Cos 0₂ Jir: 62 Cos 02 mg Hence, T₁ = sin Q₁ + Cos 0₁, tanAz = mg 8x9.8 Hence Ti = Sin 40° + cos 40⁰ tan 50° 77 SO.39455 N Cos 40° X 12 = 50.39455 x COS 450 cos so° -T₂ = 60.05788 N AL 162 a) Now given 0₁ = 40⁰ and £₂ = 50° and m= Free Body Diagram 314 01 mg To cos d₂ [Bj T2 T₂ R 0₂ 0101 = 8kg Tr my b) From A ABC sing₂ = I yo Apply Newtons second law T₂ cos 6₂ = man Ti Cost1 - Tz costz Tisins + T₂ SinG - nug = may Now l=0.3m T₁ = Sinfi + T2 sin 02 Ii Cose 1 DC = Sin 40° BC =) 0-3 = sin 40° => DC = 0.19284 1 BC=0.3m The tension for wire I is T₁ = 50.39.N and wire #2 its tension is T2 = 60.06 N b): The decoration is suspended 0.19 meter below the ceiling
A 8 kg decoration hangs from the ceiling. It is suspended by 2 wires. The first wire forms a 40° angle with the ceiling. The second wire forms a 50° angle with the ceiling. a. Find the tension in the 2 wires (including magnitude and direction) b. Assuming the wire that forms the 40° angle with the ceiling has a length of 0.3 metres, how far below the ceiling is the decoration suspended?
Transcribed Image Text:A 8 kg decoration hangs from the ceiling. It is suspended by 2 wires. The first wire forms a 40° angle with the ceiling. The second wire forms a 50° angle with the ceiling. a. Find the tension in the 2 wires (including magnitude and direction) b. Assuming the wire that forms the 40° angle with the ceiling has a length of 0.3 metres, how far below the ceiling is the decoration suspended?
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