The resultant force.

Question

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Answer 1

Question

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

(a)

Expert Solution

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:

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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)

To determine

The pitch of the wrench.

(b)

Expert Solution

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)

To determine

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

(c)

Expert Solution

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

Question

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

(a)

Expert Solution

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:

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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)

To determine

The pitch of the wrench.

(b)

Expert Solution

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)

To determine

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

(c)

Expert Solution

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

Question

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

(a)

Expert Solution

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:

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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)

To determine

The pitch of the wrench.

(b)

Expert Solution

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)

To determine

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

(c)

Expert Solution

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

Question

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

(a)

Expert Solution

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:

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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)

To determine

The pitch of the wrench.

(b)

Expert Solution

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)

To determine

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

(c)

Expert Solution

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

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

(a)

Expert Solution

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:

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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)

To determine

The pitch of the wrench.

(b)

Expert Solution

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)

To determine

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

(c)

Expert Solution

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 6

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

(a)

Expert Solution

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:

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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 7

Chapter 3.4, Problem 3.140P

(a)

To determine

The resultant force.

Answer 8

(a)

Expert Solution

Answer to Problem 3.140P

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

Explanation of Solution

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

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS W/CON >B, 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 13

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

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 14

(b)

To determine

The pitch of the wrench.

Answer 15

(b)

Expert Solution

Answer to Problem 3.140P

The pitch of the wrench is −0.0988a_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 20

(c)

To determine

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

(c)

Expert Solution

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 21

(c)

To determine

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

Answer 22

(c)

Expert Solution

Answer to Problem 3.140P

The 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

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 27

Ch. 3.1 - A foot valve for a pneumatic system is hinged at...Ch. 3.1 - 3.2A foot valve for a pneumatic system is hinged...Ch. 3.1 - It is known that a vertical force of 200 lb is...Ch. 3.1 - A 300-N force is applied at A as shown. Determine...Ch. 3.1 - A 300-N force is applied at A as shown. Determine...Ch. 3.1 - An 8-lb force P is applied to a shift lever....Ch. 3.1 - For the shift lever shown, determine the magnitude...Ch. 3.1 - An 11-lb force P is applied to a shift lever. The...Ch. 3.1 - Rod AB is held in place by the cord AC. Knowing...Ch. 3.1 - Rod AB is held in place by the cord AC. Knowing...

The resultant force.

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

Videos

The resultant force.

Answer to Problem 3.140P

Explanation of Solution

The pitch of the wrench.

Answer to Problem 3.140P

Explanation of Solution

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

Answer to Problem 3.140P

Explanation of Solution

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