Chapter 7, Problem 7.1P

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Answer 1

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Chapter 7, Problem 7.1P

Expert Solution & Answer

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

X¯=2.68ft. .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.

Now we have,

X¯=∑ wyWX¯=(w1×y1)−( w 2 × y 2 )WX¯=( 1766.5×2.5)−( 264.94×1.5)1501.56X¯=2.68ft.

Conclusion:

Therefore, the location of center of gravity of the casting is X¯=2.68ft. .

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Answer 2

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Chapter 7, Problem 7.1P

Expert Solution & Answer

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

X¯=2.68ft. .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.

Now we have,

X¯=∑ wyWX¯=(w1×y1)−( w 2 × y 2 )WX¯=( 1766.5×2.5)−( 264.94×1.5)1501.56X¯=2.68ft.

Conclusion:

Therefore, the location of center of gravity of the casting is X¯=2.68ft. .

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Answer 3

Question

100%

Chapter 7, Problem 7.1P

Expert Solution & Answer

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

X¯=2.68ft. .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.

Now we have,

X¯=∑ wyWX¯=(w1×y1)−( w 2 × y 2 )WX¯=( 1766.5×2.5)−( 264.94×1.5)1501.56X¯=2.68ft.

Conclusion:

Therefore, the location of center of gravity of the casting is X¯=2.68ft. .

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Answer 4

Question

100%

Chapter 7, Problem 7.1P

Expert Solution & Answer

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

X¯=2.68ft. .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.

Now we have,

X¯=∑ wyWX¯=(w1×y1)−( w 2 × y 2 )WX¯=( 1766.5×2.5)−( 264.94×1.5)1501.56X¯=2.68ft.

Conclusion:

Therefore, the location of center of gravity of the casting is X¯=2.68ft. .

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Answer 5

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Answer 6

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Answer 7

Expert Solution & Answer

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

X¯=2.68ft. .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.

Now we have,

X¯=∑ wyWX¯=(w1×y1)−( w 2 × y 2 )WX¯=( 1766.5×2.5)−( 264.94×1.5)1501.56X¯=2.68ft.

Conclusion:

Therefore, the location of center of gravity of the casting is X¯=2.68ft. .

Answer 8

Expert Solution & Answer

To determine

The location of the center of gravity of the casting.

Answer to Problem 7.1P

X¯=2.68ft. .

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.

Now we have,

X¯=∑ wyWX¯=(w1×y1)−( w 2 × y 2 )WX¯=( 1766.5×2.5)−( 264.94×1.5)1501.56X¯=2.68ft.

Conclusion:

Therefore, the location of center of gravity of the casting is X¯=2.68ft. .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The location of the center of gravity of the casting.

Answer 12

Answer to Problem 7.1P

X¯=2.68ft. .

Answer 13

Explanation of Solution

Given information:

Given a cylindrical cast iron casting having an axial hole extending pathway through the casting.

Applied Statics and Strength of Materials (6th Edition), Chapter 7, Problem 7.1P

The centroid of a body can be found with two axes taken into consideration. As, the given figure is symmetrical about X-axis, so the centroid must lie on the X-axis. Now, by calculating the weight of the given member with reference to densities taken from U S Customary codes for typical average properties of some common materials. Thereby, in reference to the codes the weight of cast iron is taken as 450pcf .

Calculation:

Let W₁the weight of the whole cylinder and W₂ the weight of the empty space considering its density same as that of the cast iron.

w1=πr2h×450pcfw1=3.14×612×612×5×450w1=1766.5lb.w2 =πr2h×450pcfw2 =3.14×312×312×3×450w2 =264.94lb.W=w1+w2W=1766.5lb+(−264.94)lbW=1502.56lb.