(a) Replace the force F = − 2800 i + 1600 j + 3000 k lb acting at end A of the crank handle with a force R acting at O and a couple- vector C R . (b) Resolve R into the normal component P (normal to the cross section of the shaft) and the shear component V (in the plane of the cross section). (c) Resolve C R into the twisting component T and the bending component M .

Question

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Answer 1

Textbook Question

Chapter 2, Problem 2.91P

(a) Replace the force F = − 2800 i + 1600 j + 3000 k lb acting at end A of the crank handle with a force R acting at O and a couple-vector C R . (b) Resolve R into the normal component P (normal to the cross section of the shaft) and the shear component V (in the plane of the cross section). (c) Resolve C R into the twisting component T and the bending component M.

Chapter 2, Problem 2.91P, (a) Replace the force F=2800i+1600j+3000klb acting at end A of the crank handle with a force R

Expert Solution

To determine

(i)

The equivalent force couple vector acting at O.

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Expert Solution

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Expert Solution

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in

Conclusion:

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in.

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Answer 2

Textbook Question

Chapter 2, Problem 2.91P

(a) Replace the force F = − 2800 i + 1600 j + 3000 k lb acting at end A of the crank handle with a force R acting at O and a couple-vector C R . (b) Resolve R into the normal component P (normal to the cross section of the shaft) and the shear component V (in the plane of the cross section). (c) Resolve C R into the twisting component T and the bending component M.

Chapter 2, Problem 2.91P, (a) Replace the force F=2800i+1600j+3000klb acting at end A of the crank handle with a force R

Expert Solution

To determine

(i)

The equivalent force couple vector acting at O.

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Expert Solution

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Expert Solution

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in

Conclusion:

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in.

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Answer 3

Textbook Question

Chapter 2, Problem 2.91P

(a) Replace the force F = − 2800 i + 1600 j + 3000 k lb acting at end A of the crank handle with a force R acting at O and a couple-vector C R . (b) Resolve R into the normal component P (normal to the cross section of the shaft) and the shear component V (in the plane of the cross section). (c) Resolve C R into the twisting component T and the bending component M.

Chapter 2, Problem 2.91P, (a) Replace the force F=2800i+1600j+3000klb acting at end A of the crank handle with a force R

Expert Solution

To determine

(i)

The equivalent force couple vector acting at O.

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Expert Solution

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Expert Solution

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in

Conclusion:

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in.

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Answer 4

Textbook Question

Chapter 2, Problem 2.91P

(a) Replace the force F = − 2800 i + 1600 j + 3000 k lb acting at end A of the crank handle with a force R acting at O and a couple-vector C R . (b) Resolve R into the normal component P (normal to the cross section of the shaft) and the shear component V (in the plane of the cross section). (c) Resolve C R into the twisting component T and the bending component M.

Chapter 2, Problem 2.91P, (a) Replace the force F=2800i+1600j+3000klb acting at end A of the crank handle with a force R

Expert Solution

To determine

(i)

The equivalent force couple vector acting at O.

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Expert Solution

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Expert Solution

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in

Conclusion:

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(i)

The equivalent force couple vector acting at O.

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Expert Solution

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Expert Solution

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in

Conclusion:

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in.

Answer 8

Expert Solution

To determine

(i)

The equivalent force couple vector acting at O.

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(i)

The equivalent force couple vector acting at O.

Answer 13

Answer to Problem 2.91P

The magnitude of force at O is−2800i^+1600j^+3000k^ lb.

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Answer 14

Explanation of Solution

Given Information:

F=-2800i+1600j+3000k

Concept used:

When a force is moved by a distance of x, the resulting system willhave the same magnitude of the force and the moment(equal to the force×distance) associated to that pointC=M=F*distance by which the force is moved.

When a force is moved the magnitude of the force does not change,F=R=−2800i^+1600j^+3000k^ lbTaking anti−clockwise moment as a positive moment.The couple acting at O is:CR=rOA×FrOA=10i^+5j^−4k^CR=rOA×FCR=|( i j k 10 5 −4 −2800 1600 3000 )|CR=i(5*3000−(−4*1600))−j(10*3000−(−4*−2800))+k(10*1600−(5*−2800))CR=−21400i^−18800j^+30000k^ lb−in

Conclusion:

The magnitude of couple at O is−21400i^−18800j^+30000k^ lb−in.

Answer 15

Expert Solution

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(ii)

The resolved components of force R into normal component P and the shear component V.

Answer 20

Answer to Problem 2.91P

The normal component is 1600 lb.

The shear component is 4100 lb.

Answer 21

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The normal component=RyShear component=Rx2+Ry2

Calculation:

The normal component=Ry=1600 lbShear component=Rx2+Ry2Shear component= ( −2800 )2+ 16002Shear component=4100 lb

Conclusion:

The normal component is 1600 lb.

The shear component is 4100 lb.

Answer 22

Expert Solution

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in

Conclusion:

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

(iii)

The resolved components of Resultant couple C^R into Torque component T and the bending moment component M.

Answer 27

Answer to Problem 2.91P

The torque component is 18800 lb-in.

The bending moment component is 36900 lb-in

Answer 28

Explanation of Solution

Given:

F=-2800i+1600j+3000k

Concept Used:

The Torque component=CRyThe bending moment component= ( C x R )2+ ( C z R )2

Calculation:

The Torque component=CRy⇒18800 lb−inThe bending moment component= ( C x R )2+ ( C z R )2The bending moment component= ( 21400 )2+ ( 30000 )2⇒36900 lb−in