Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 5.4, Problem 5.132P

Chapter 5.4, Problem 5.132P, PROBLEM 5.132 The sides and the base of a punch bowl are of uniform thickness t. If t  R and R = 250

PROBLEM 5.132

The sides and the base of a punch bowl are of uniform thickness t. If t << R and R = 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the bunch.

(a)

Expert Solution
Check Mark
To determine

The location of the centre of gravity of the bowl.

Answer to Problem 5.132P

The location of the centre of gravity of the bowl is x¯=0, y¯=121.9mm and z¯=12R.

Explanation of Solution

Refer Fig. P5.132 and Fig. 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 5.4, Problem 5.132P , additional homework tip  1

For the coordinate axis given below, using the symmetry of the diagram, determine the x and z coordinates of the centre of gravity.

x¯=0z¯=0

The bowl can be assumed as a shell, where the centre of gravity coincides with the centroid of the shell.

The element of area is obtained by rotating the arc ds about the y-axis for the walls of the bowl.

Write the expression for the area of the element.

dAwall=(2πRsinθ)(Rdθ) (I)

Here, dAwall is the area of the element and R is the radius of the shell.

Write the expression for y coordinate of the centroid of the element.

(y¯EL)wall=Rcosθ (II)

Here, y¯EL is the y coordinate of the centroid of the element.

Write the expression for y¯wallAwall of the element.

y¯wallAwall=(y¯EL)walldA (III)

Write the expression for y coordinate of the centre of gravity.

y¯=Σy¯AΣA (IV)

Here, Σy¯A is the sum of the product of the y coordinate and the area and ΣA is the sum of the areas.

Conclusion:

Calculate the area using equation (I).

Awall=π/6π/2dA=π/6π/22πR2sinθdθ=2πR2[cosθ]π/6π/2=π3R2 (V)

Substitute (I) and (II) in (III).

y¯wallAwall=π/6π/2(Rcosθ)(2πR2sinθdθ)=πR3[cos2θ]π/6π/2=34πR3 (VI)

From figure 1, find the area of the base and distance of the centroid of the base to the y axis.

Abase=π4R2y¯base=32R (VII)

Substitute equations (V), (VI) and (VII) in equation (IV).

y¯=(34πR3)+(π4R2)(32R)π3R2+π4R2=0.48763R (VIII)

Substitute 250mm for R in equation (VIII) to find the y coordinate of the centre of gravity.

y¯=(0.48763)(250mm)=121.9mm

Therefore, the location of the centre of gravity of the bowl is x¯=0, y¯=121.9mm and z¯=12R.

(b)

Expert Solution
Check Mark
To determine

The location of the centre of gravity of the punch.

Answer to Problem 5.132P

The location of the centre of gravity of the bowl is x¯=0, y¯=90.2mm and z¯=12R.

Explanation of Solution

Refer Fig. P5.132 and Fig. 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 5.4, Problem 5.132P , additional homework tip  2

For the coordinate axis given below, using the symmetry of the diagram, determine the x and z coordinates of the centre of gravity.

x¯=0z¯=0

The punch can be assumed as homogenous, where the centre of gravity coincides with the centroid of the volume.

The element of volume of the disk has radius x and thickness dy.

Find the expression for x2 of the element from the equation of the curve.

x2+y2=R2x2=R2y2

Write the expression for the volume of the element.

dV=πx2dy=π(R2y2)dy (IX)

Here, R is the radius of the punch, and y is distance of the element from the top.

Write the expression for y coordinate of the centre of gravity.

y¯=y¯ELdVV (X)

Conclusion:

Calculate the volume using equation (IX).

V=3/2R0dV=3/2R0π(R2y2)dy=π[R2y13y3]3/2R0=π[R2(32R)13(32R)3]=38π3R3 (XI)

Find y¯ELdV.

y¯ELdV=3/2R0y[π(R2y2)dy]=π[12R2y214y4]3/2R0=π[12R2(32R)214(32R)4]=1564πR4 (XII)

Substitute equations (XI), and (XII) in equation (X).

y¯=1564πR438π3R3=583R (XIII)

Substitute 250mm for R in equation (XIII) to find the y coordinate of the centre of gravity.

y¯=583(250mm)=90.2mm

Therefore, the location of the centre of gravity of the bowl is x¯=0, y¯=90.2mm and z¯=12R.

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Chapter 5 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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