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Parents Pricing Home NSC Mechanical Technology Fitting and Machining Basic calculations on forces, moments and stress and strain FIGUUR 8.1 toon 'n stel van drie kragte met vier kragte wat op dieselfde punt inwerk
FIGUUR 8.1 toon 'n stel van drie kragte met vier kragte wat op dieselfde punt inwerk - NSC Mechanical Technology Fitting and Machining - Question 8 - 2021 - Paper 1 Question 8
View full question FIGUUR 8.1 toon 'n stel van drie kragte met vier kragte wat op dieselfde punt inwerk.
Bereken die volgende:
8.1.1 Som van die horizontale komponente (HK)
8.1.2 So... show full transcript
View marking scheme Worked Solution & Example Answer:FIGUUR 8.1 toon 'n stel van drie kragte met vier kragte wat op dieselfde punt inwerk - NSC Mechanical Technology Fitting and Machining - Question 8 - 2021 - Paper 1
Som van die horizontale komponente (HK) Only available for registered users.
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Om die som van die horizontale komponente te bereken, gebruik ons die volgende formules:
e x t S o m e x t H C = 280 i m e s e x t c o s ( 4 5 e x t o ) + 120 i m e s e x t c o s ( 0 e x t o ) − 150 i m e s e x t c o s ( 9 0 e x t o ) − 250 i m e s e x t c o s ( 3 0 e x t o ) ext{Som} ext{HC} = 280 imes ext{cos}(45^ ext{o}) + 120 imes ext{cos}(0^ ext{o}) - 150 imes ext{cos}(90^ ext{o}) - 250 imes ext{cos}(30^ ext{o}) e x t S o m e x t H C = 280 im ese x t cos ( 4 5 e x t o ) + 120 im ese x t cos ( 0 e x t o ) − 150 im ese x t cos ( 9 0 e x t o ) − 250 im ese x t cos ( 3 0 e x t o ) Na berekening verkry ons:
e x t S o m e x t H C = 101 , 48 e x t N . ext{Som} ext{HC} = 101,48 ext{ N}. e x t S o m e x t H C = 101 , 48 e x t N .
Som van die vertikale komponente (VK) Only available for registered users.
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Die som van die vertikale komponente word bereken met die volgende formules:
e x t S o m e x t V C = 280 i m e s e x t s i n ( 4 5 e x t o ) + 120 i m e s e x t s i n ( 0 e x t o ) − 150 i m e s e x t s i n ( 9 0 e x t o ) − 250 i m e s e x t s i n ( 3 0 e x t o ) ext{Som} ext{VC} = 280 imes ext{sin}(45^ ext{o}) + 120 imes ext{sin}(0^ ext{o}) - 150 imes ext{sin}(90^ ext{o}) - 250 imes ext{sin}(30^ ext{o}) e x t S o m e x t V C = 280 im ese x t s in ( 4 5 e x t o ) + 120 im ese x t s in ( 0 e x t o ) − 150 im ese x t s in ( 9 0 e x t o ) − 250 im ese x t s in ( 3 0 e x t o ) Na die berekening kom ons by:
e x t S o m e x t V C = − 77 , 01 e x t N . ext{Som} ext{VC} = -77,01 ext{ N}. e x t S o m e x t V C = − 77 , 01 e x t N .
Grootte van die ewewigsrag Only available for registered users.
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Die grootte van die ewewigsrag word bereken deur:
E = e x t s q r t ( e x t V C 2 + e x t H C 2 ) E = ext{sqrt}( ext{VC}^2 + ext{HC}^2) E = e x t s q r t ( e x t V C 2 + e x t H C 2 ) Hierna bereken ons:
E = e x t s q r t ( ( − 77 , 01 ) 2 + ( 101 , 48 ) 2 ) i g h t a r r o w E = 127 , 39 e x t N . E = ext{sqrt}((-77,01)^2 + (101,48)^2)
ightarrow E = 127,39 ext{ N}. E = e x t s q r t (( − 77 , 01 ) 2 + ( 101 , 48 ) 2 ) i g h t a rro wE = 127 , 39 e x t N .
Rigting van die ewewigsrag Only available for registered users.
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Om die rigting van die ewewigsrag te vind, gebruik ons:
an( heta) = rac{ ext{VC}}{ ext{HC}}
ightarrow heta = an^{-1}igg( rac{-77,01}{101,48}igg)
Die rigting is 37,19° N van W.
Bereken die groote van die reaksies by stutte A en B. Only available for registered users.
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Vir die reaksies by stutte A en B gebruik ons die momentvergelyking:
e x t S o m v a n d i e h o r i s o n t a l e k r a g t e = 0 e x t e n e x t S o m v a n d i e v e r t i k a l e k r a g t e = 0 ext{Som van die horisontale kragte = 0} ext{ en } ext{Som van die vertikale kragte = 0} e x t S o m v an d i e h or i so n t a l e k r a g t e = 0 e x t e n e x t S o m v an d i e v er t ika l e k r a g t e = 0 Hierna bereken ons die noodsaaklike reaksies en vind ons dat:
Reaksie A = 554,25 N en Reaksie B = (af te lei vanaf die balans van die kragte).
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