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#4 10₁ = 40° 0₂ = 50° and wire #2 - 11 TIL Ti Sindi putting 1₂ O₂ Ti Cos T₂ cos 0₂ => T₁ Cos 0₁ = T₂ Cos 0₂ > T2₂ = Ti Costi COS 0₂ 0- and by Sin e T₁- = 8kg/m 1 m = Ti sin + + 0₁ (wire #1 - Decoration Diagram. + T₂ Sin02 = 6 T1= 42.735 N we are going to use equation latel cos i Cos 0₂ 8x9.8 T₁ = sin 40° + cos 40° Cos 500 mg Sin 01 + cos01 Cos 02 mg Ti Costi Cos 0₂ = mg 01₂ 42.735 * cos 50° Cos 400 X T2 = 46.297 N mg Free Body Diagram T 0₂ 72 F U 1 12 cos D₂ A 1 Ti 1 0₂ Nor B my 01 /C Apply Newtons second law TI Cos DI T₂ cos 62 = Mau Ti sin 1 + 1₂ sing-mg = may 2 AB b) From A ABC sing₁= I AB= lsing l=0.3m T₁=SnQ₁ + T2 sin 02 Ji Costi 0.3x sin 40° AB= 0.193 m a): The tension for wire I is T₁ = 42.735 N and wire #2 its tension is T₂ = 46.297 N b). The distance for how fa the ceiling is, where the decoration is suspended is AB = 0.193 m

#4 10₁ = 40° 0₂ = 50° and wire #2 - 11 TIL Ti Sindi putting 1₂ O₂ Ti Cos T₂ cos 0₂ => T₁ Cos 0₁ = T₂ Cos 0₂ > T2₂ = Ti Costi COS 0₂ 0- and by Sin e T₁- = 8kg/m 1 m = Ti sin + + 0₁ (wire #1 - Decoration Diagram. + T₂ Sin02 = 6 T1= 42.735 N we are going to use equation latel cos i Cos 0₂ 8x9.8 T₁ = sin 40° + cos 40° Cos 500 mg Sin 01 + cos01 Cos 02 mg Ti Costi Cos 0₂ = mg 01₂ 42.735 * cos 50° Cos 400 X T2 = 46.297 N mg Free Body Diagram T 0₂ 72 F U 1 12 cos D₂ A 1 Ti 1 0₂ Nor B my 01 /C Apply Newtons second law TI Cos DI T₂ cos 62 = Mau Ti sin 1 + 1₂ sing-mg = may 2 AB b) From A ABC sing₁= I AB= lsing l=0.3m T₁=SnQ₁ + T2 sin 02 Ji Costi 0.3x sin 40° AB= 0.193 m a): The tension for wire I is T₁ = 42.735 N and wire #2 its tension is T₂ = 46.297 N b). The distance for how fa the ceiling is, where the decoration is suspended is AB = 0.193 m

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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?
#4 wire # 2. 0₁ = 40° 0₂ = 50° 10₂ 01 a) Ti Cos 0₁ - T₂ cos A₂ = G F 27 T₁ Cos 01 = T₂ Cos 0₂ > T2 Ti Costi COS 02 → Ti Ti = and Ti Sin 0₁ + T₂ Sinf₂ = • mg putting T₂ [8 kg] T₁ Sindi Ti sini + 8₁ (wire #1 - D Decoration Diagram 8x9.8 Sin 400 + we are going to use equation later cos er Cos 0₂ mg Sin 01 + coS01 c0382 T₁ = 42, 735 N + cos 40° Cos 500 Ti Costi Cos 0₂ - mg Cos 400 0 = and by 12 42.735 * cos 50° T₂ = 46.297 N X • mg Free Body Diagram 2010 Ti 202 T2 0₂ N101 I cos 0₂ [Bj my > l=0.3m T₁ = Sin Q₁ + T2 sin 02 II COSE I Apply Newtons second law TI Cos 01 - T₂ cos 6₂ = mau Ti sin 01 + T₂ Sing-mg AB 6) From A ABC sing, I AB= lsing₁ = 0.3x sin 40° AB= 0.193 m. = may The tension for wire 1 is T₁ = 42.735 N and wire #2 its tension is 7₂ = 46.297 N X b) The distance for how far the ceiling is where the decoration is suspended is AB = 0.193 m
Transcribed Image Text:#4 wire # 2. 0₁ = 40° 0₂ = 50° 10₂ 01 a) Ti Cos 0₁ - T₂ cos A₂ = G F 27 T₁ Cos 01 = T₂ Cos 0₂ > T2 Ti Costi COS 02 → Ti Ti = and Ti Sin 0₁ + T₂ Sinf₂ = • mg putting T₂ [8 kg] T₁ Sindi Ti sini + 8₁ (wire #1 - D Decoration Diagram 8x9.8 Sin 400 + we are going to use equation later cos er Cos 0₂ mg Sin 01 + coS01 c0382 T₁ = 42, 735 N + cos 40° Cos 500 Ti Costi Cos 0₂ - mg Cos 400 0 = and by 12 42.735 * cos 50° T₂ = 46.297 N X • mg Free Body Diagram 2010 Ti 202 T2 0₂ N101 I cos 0₂ [Bj my > l=0.3m T₁ = Sin Q₁ + T2 sin 02 II COSE I Apply Newtons second law TI Cos 01 - T₂ cos 6₂ = mau Ti sin 01 + T₂ Sing-mg AB 6) From A ABC sing, I AB= lsing₁ = 0.3x sin 40° AB= 0.193 m. = may The tension for wire 1 is T₁ = 42.735 N and wire #2 its tension is 7₂ = 46.297 N X b) The distance for how far the ceiling is where the decoration is suspended is AB = 0.193 m
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