A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine ( a ) the resultant force R, ( b ) the pitch of the wrench, ( c ) the point where the axis of the wrench intersects the xz -plane.

Question

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Answer 1

Textbook Question

Chapter 3.4, Problem 3.140P

A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

Chapter 3.4, Problem 3.140P, A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P,

(a)

Expert Solution

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

(b)

Expert Solution

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

(c)

Expert Solution

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

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Answer 2

Textbook Question

Chapter 3.4, Problem 3.140P

A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

Chapter 3.4, Problem 3.140P, A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P,

(a)

Expert Solution

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

(b)

Expert Solution

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

(c)

Expert Solution

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

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Answer 3

Textbook Question

Chapter 3.4, Problem 3.140P

A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

Chapter 3.4, Problem 3.140P, A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P,

(a)

Expert Solution

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

(b)

Expert Solution

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

(c)

Expert Solution

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

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Answer 4

Textbook Question

Chapter 3.4, Problem 3.140P

A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

Chapter 3.4, Problem 3.140P, A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P,

(a)

Expert Solution

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

(b)

Expert Solution

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

(c)

Expert Solution

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

(b)

Expert Solution

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

(c)

Expert Solution

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Answer 8

(a)

Expert Solution

To determine

The resultant force.

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The resultant force.

Answer 13

Answer to Problem 3.140P

The resultant force is (3P/25)[2i^−20j^−k^]_.

Answer 14

Explanation of Solution

The diagram for the force-couple system is given below:

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 3.4, Problem 3.140P

Refer fig 1.

Write the equation of resultant force.

R→=PλBA+PλDC+PλDE=P[λBA+λDC+λDE] (I)

Here, the resultant force is R→, the magnitude of the forces are P, the direction of the force at points is λBA,λDC,λDE

Conclusion:

Substitute, (45j^−35k^) for λBA, (35i^−45j^) for λDC, (35i^−45j^) for λDE in equation (I)

R→=P[(45j^−35k^)+(35i^−45j^)+(35i^−45j^)]=3P25[2i^−20j^−k^]

The magnitude of the resultant force,

|R→|=(3P/25)2[(2)2+(20)2+(1)2]=(275/25)P

Thus, the resultant force is (3P/25)[2i^−20j^−k^]_.

Answer 15

(b)

Expert Solution

To determine

The pitch of the wrench.

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The pitch of the wrench.

Answer 20

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Answer 21

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (II)

Here, the pitch of the wrench is p, the momentum along wrench is M1

Since, the R→ and M→oR are not in same direction, so the form of the wrench is,

M→1=λaxisM→oR (III)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=(r→1×F→1)+(r→2×F→2)+(r→3×F→3) (IV)

Here, the momentum is M→BR,M→A,M→B, the distance is r1.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→1×F→1)+(r→2×F→2)+(r→3×F→3)] (V)

Conclusion:

Substitute, (3P/25)[2i^−20j^−k^] for R→, (275/25)P for R, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (V).

M→1=(3P/25)[2i^−20j^−k^](275/25)P[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=(−8Pa675)(2i^−20j^−k^)=(−8Pa155)

Substitute, (−8Pa155) for M1, (275/25)P for R in equation (II)

p=(−8Pa/155)(275/25)P=−0.0988a

Thus, the pitch of the wrench is −0.0988a_.

Answer 22

(c)

Expert Solution

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The point at which the axis of wrench intersects the xz-plane.

Answer 27

Answer to Problem 3.140P

The axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Answer 28

Explanation of Solution

Refer fig 1,

Write the equation for the force couple system for the wrench.

M→BR=M→1+M→2 (VI)

Here, the momentum is M→2.

Write the expression for the momentum at which the wrench intersects the xz-plane.

M→2=r→Q/O×R→ (VII)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=xi^+zk^

Here, the coordinates are x,z.

Conclusion:

Substitute, (24a)j^ for r→1, (−4P5−3P5)k^ for F→1, (20a)j^ for r→2, (3P5i^−4P5j^) for F→2, (20a)j^ for r→3, (−9P25i^−4P5j^+12P25k^) for F→3 in equation (IV).

M→oR=[{(24a)j^×(−4P5−3P5)k^}+{(20a)j^×(3P5i^−4P5j^)}+{(20a)j^×(−9P25i^−4P5j^+12P25k^)}]=[(24Pa5)(−i^−k^)]

Substitute, [(24Pa5)(−i^−k^)] for M→oR, (−8Pa675)(2i^−20j^−k^) for M→1 in equation (VI).

[(24Pa5)(−i^−k^)]=[(−8Pa675)(2i^−20j^−k^)]+M→2M→2={(8Pa675)[−430i^−20j^−406k^]}

Substitute, {(8Pa675)[−430i^−20j^−406k^]} for M→2, {(3P25)[2i^+20j^−k^]} for R→, (xi^+zk^) for r→Q/O in equation (VII).

{(8Pa675)[−430i^−20j^−406k^]}=(xi^+zk^)×{(3P25)[2i^+20j^−k^]}

Comparing the coefficients of the x and z components both sides,

x=2a, z=−1.99a

Therefore, he axis of wrench intersects the xz-plane at x=2a, z=−1.99a_.

Answer 29

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Solution Summary: The diagram for the force-couple system is given below. Write the equation of resultant force.

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Answer to Problem 3.140P

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Answer to Problem 3.140P

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Answer to Problem 3.140P

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