Mathematics for Machine Technology
Mathematics for Machine Technology
7th Edition
ISBN: 9781133281450
Author: John C. Peterson, Robert D. Smith
Publisher: Cengage Learning
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Textbook Question
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Chapter 62, Problem 28A

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.
Chapter 62, Problem 28A, Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer

Expert Solution & Answer
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To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3V4V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5cm for a and 2.5cm for h1 in Equation (II).

V1=(6.5cm)2(2.5cm)=(6.5cm)(6.5cm)(2.5cm)=105.63cm3

Substitute 6.5cm for l2, 1cm for b2 and 3.5cm for h2 in Equation (III).

V2=(6.5cm)(1cm)(3.5cm)=(6.5cm)(3.5 cm2)=22.75cm3

Substitute 6.5cm for l3, 1.5cm for b3 and 1cm for h3 in Equation (IV).

V3=(6.5cm)(1.5cm)(1cm)=(6.5cm)(1.5 cm2)=9.75cm3

Substitute 1.5cm for r and 1cm for t4 in Equation (V).

V4=π(1.5cm)2(1cm)=π(2.25 cm2)(1cm)=7.07cm3

Substitute 2.5cm for b5, 2cm for h5 and 1cm for t5 in Equation (VI).

V5=12(2.5cm)(2cm)(1cm)=(0.5)(2.5cm)(2cm)(1cm)=2.5cm3

Substitute 105.63cm3 for V1, 2.5cm3 for V5, 7.07cm3 for V4, 9.75cm3 for V3 and 22.75cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)(7.07 cm3)(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)(7.07 cm3)(2.5 cm3)=151.31cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31cm3.

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