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Parents Pricing Home NSC Mechanical Technology Welding and Metalwork Developments and templates FIGURE 11.1 shows a square-to-square off-centre hopper with a vertical height of 400 mm
FIGURE 11.1 shows a square-to-square off-centre hopper with a vertical height of 400 mm - NSC Mechanical Technology Welding and Metalwork - Question 11 - 2021 - Paper 1 Question 11
View full question FIGURE 11.1 shows a square-to-square off-centre hopper with a vertical height of 400 mm. Answer the questions that follow.
Calculate the true lengths of the followi... show full transcript
View marking scheme Worked Solution & Example Answer:FIGURE 11.1 shows a square-to-square off-centre hopper with a vertical height of 400 mm - NSC Mechanical Technology Welding and Metalwork - Question 11 - 2021 - Paper 1
A–2 Only available for registered users.
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To calculate the true length A–2, we use the formula for calculating the true length between two points in 3D geometry:
e x t T r u e l e n g t h ( A − 2 ) = e x t s q r t ( ( 400 ) 2 + ( 280 ) 2 + ( 400 ) 2 ) = e x t s q r t ( 576000 + 78400 + 160000 ) = e x t s q r t ( 296000 ) e x t T r u e l e n g t h ( A − 2 ) e x t i s a p p r o x i m a t e l y 544.06 e x t m m ( o r 544 m m ) ext{True length (A - 2)} = ext{sqrt}((400)^2 + (280)^2 + (400)^2)
= ext{sqrt}(576000 + 78400 + 160000)
= ext{sqrt}(296000)
\\
ext{True length (A - 2)} ext{ is approximately } 544.06 ext{ mm (or 544 mm)} e x t T r u e l e n g t h ( A − 2 ) = e x t s q r t (( 400 ) 2 + ( 280 ) 2 + ( 400 ) 2 ) = e x t s q r t ( 576000 + 78400 + 160000 ) = e x t s q r t ( 296000 ) e x t T r u e l e n g t h ( A − 2 ) e x t i s a pp ro x ima t e l y 544.06 e x t mm ( or 544 mm )
C–3 Only available for registered users.
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To find the true length C–3, we apply the same method:
e x t T r u e l e n g t h ( C − 3 ) = e x t s q r t ( ( 220 ) 2 + ( 60 ) 2 + ( 400 ) 2 ) = e x t s q r t ( 48400 + 3600 + 160000 ) = e x t s q r t ( 212000 ) e x t T r u e l e n g t h ( C − 3 ) e x t i s a b o u t 460.43 e x t m m ( o r 460 m m ) ext{True length (C - 3)} = ext{sqrt}((220)^2 + (60)^2 + (400)^2)
= ext{sqrt}(48400 + 3600 + 160000)
= ext{sqrt}(212000)
\\
ext{True length (C - 3)} ext{ is about } 460.43 ext{ mm (or 460 mm)} e x t T r u e l e n g t h ( C − 3 ) = e x t s q r t (( 220 ) 2 + ( 60 ) 2 + ( 400 ) 2 ) = e x t s q r t ( 48400 + 3600 + 160000 ) = e x t s q r t ( 212000 ) e x t T r u e l e n g t h ( C − 3 ) e x t i s ab o u t 460.43 e x t mm ( or 460 mm )
A–B Only available for registered users.
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For the truncated cone, we calculate the true length A–B using the formula:
ext{True length (A - B)} = rac{ ext{D}}{12} \\
= rac{ uxtimes600}{12}
= rac{1884.96}{12} = 157.08 ext{ mm (approximately 157 mm)}
Circumference of the top circle Only available for registered users.
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The circumference of the top circle is given by the formula:
e x t C i r c u m f e r e n c e = e x t π i m e s D = e x t π i m e s 400 e x t T h e r e f o r e , C i r c u m f e r e n c e i s a p p r o x i m a t e l y 1256.64 e x t m m ( o r 1257 m m ) ext{Circumference} = ext{π} imes D \\
= ext{π} imes 400 \\
ext{Therefore, Circumference is approximately } 1256.64 ext{ mm (or 1257 mm)} e x t C i rc u m f ere n ce = e x t π im esD = e x t π im es 400 e x t T h ere f ore , C i rc u m f ere n ce i s a pp ro x ima t e l y 1256.64 e x t mm ( or 1257 mm )
True vertical height of the truncated cone Only available for registered users.
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The true vertical height of the truncated cone is simply the vertical height given in the problem, which is:
600 mm.
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