Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as M = 2 3 μ k P sin θ R 2 3 − R 1 3 R 2 2 − R 1 2 .

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Answer 1

Question

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .

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Answer 2

Question

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .

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Answer 3

Question

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .

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Answer 4

Question

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .

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Answer 5

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .

Answer 6

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .

Answer 7

Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=23μkPsinθR23−R13R22−R12$ .

Answer 8

Expert Solution & Answer

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=23μkPsinθR23−R13R22−R12$

Here, the coefficient of kinetic friction is $μk$ , the magnitude of the total axial force is P, the outer radii of the collar is $R1$ , and the inner radii of the collar is $R2$ .

Show the free-body diagram of the conical bearing as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 8.3, Problem 8.96P

It has been assumed that the normal force per unit area is inversely proportional to r.

$ΔA=rΔsΔϕ$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $Δr$ , and the change in azimuthal angle is $Δϕ$ .

Here, $Δs=Δrsinθ$

Find the normal force exerted $ΔN$ due to the element of area $ΔA$ using the relation;

$ΔN=kΔA$

Substitute $rΔsΔϕ$ for $ΔA$ .

$ΔN=krΔsΔϕ$

Substitute $Δrsinθ$ for $Δs$ .

$ΔN=krΔrsinθΔϕ$ (1)

Find the change in friction force in y-axis;

$ΔFy=ΔNsinθ$

Substitute $krΔrsinθΔϕ$ for $ΔN$ .

$ΔFy=krΔrsinθΔϕsinθ=krΔrΔϕ$

Find the value of k using the relation.

$P=∑ΔFy=∫02π∫R1R2ΔFy$

Substitute $krΔrΔϕ$ for $ΔFy$ .

$P=∫02π∫R1R2krΔrΔϕ=k[r22]R1R2[ϕ]02π=k[R22−R122][2π]=πk(R22−R12)$

$k=Pπ(R22−R12)$

Substitute $Pπ(R22−R12)$ for k in Equation (1).

$ΔN=Pπ(R22−R12)rΔrsinθΔϕ$

Find the change in friction force $ΔF$ using the relation.

$ΔF=μkΔN$

Substitute $Pπ(R22−R12)rΔrsinθΔϕ$ for $ΔN$ .

$ΔF=μkPπ(R22−R12)rΔrsinθΔϕ$

Find the change in moment $ΔM$ using the relation.

$ΔM=rΔF$

Substitute $μkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔF$ .

$ΔM=rμkPπ(R22−R12)rΔrsinθΔϕ$

Integrate the equation to find the couple M.

$M=∫02π∫R1R2ΔM$

Substitute $rμkPπ(R22−R12)rΔrsinθΔϕ$ for $ΔM$ and integrate.

$M=∫02π∫R1R2rμkPπ(R22−R12)rΔrsinθΔϕ=μkPπsinθ(R22−R12)[r33]R1R2[ϕ]02π=μkPπsinθ(R22−R12)[R23−R133][2π]=23μkPsinθ(R23−R13)(R22−R12)$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=23μkPsinθ(R23−R13)(R22−R12)$ .