The resultant force.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

1 8 4 Add numbers so that the sum of any row or column equals .30 Use only these numbers: .1.2.3.4.5.6.10.11.12.12.13.14.14

Uppgift 2 (9p) I77777 20 kN 10 kN/m 4 [m] 2 2 Bestäm tvärkrafts- och momentdiagram för balken i figuren ovan. Extrempunkter ska anges med både läge och värde i diagrammen.

**Problem 8-45.** The man has a mass of 60 kg and the crate has a mass of 100 kg. If the coefficient of static friction between his shoes and the ground is \( \mu_s = 0.4 \) and between the crate and the ground is \( \mu_c = 0.3 \), determine if the man is able to move the crate using the rope-and-pulley system shown. **Diagram Explanation:** The diagram illustrates a scenario where a man is attempting to pull a crate using a rope-and-pulley system. The setup is as follows: - **Crate (C):** Positioned on the ground with a rope attached. - **Rope:** Connects the crate to a pulley system and extends to the man. - **Pulley on Tree:** The rope runs over a pulley mounted on a tree which redirects the rope. - **Angles:** - The rope between the crate and tree forms a \(30^\circ\) angle with the horizontal. - The rope between the tree and the man makes a \(45^\circ\) angle with the horizontal. - **Man (A):** Pulling on the rope with the intention of moving the crate. This arrangement tests the…

Answer 2

Question

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Answer 6

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Answer 7

Chapter 3.4, Problem 3.139P

(a)

To determine

The resultant force.

Answer 8

(a)

Expert Solution

Answer to Problem 3.139P

The resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation of the distance between AC.

dAC=r12+r22+r32 (I)

Here, the distance between AC is dAC, the coordinates are r1,r2,r3

Write the equation of the distance between BD.

dBD=r42+r52+r62 (II)

Here, the distance between BD is dBD, the coordinates are r4,r5,r6

Write the equation of the momentum about AC.

T→AC=(d→ACdAC)F1 (III)

Here, the momentum about AC is T→AC, the force is F1.

Write the equation of the momentum about BD.

T→BD=(d→BDdBD)F2 (IV)

Here, the momentum about BD is T→BD, the force is F2.

Write the equation of resultant force.

R→=T→AC+T→BD (V)

Here, the resultant force is R→.

Conclusion:

Substitute, 6 m for r1, 2 m for r2, 9 m for r3 in equation (I)

dAC=(6 m)2+(2 m)2+(9 m)2=11 m

Substitute, 14 m for r4, 2 m for r5, 5 m for r5 in equation (II)

dBD=(14 m)2+(2 m)2+(5 m)2=15 m

Substitute, (6 m)i^+(2 m)j^+(9 m)k^ for d→AC, (11 m) for dAC, 1650 N for F1 in equation (III).

T→AC=([(6 m)i^+(2 m)j^+(9 m)k^](11 m))(1650 N)=[(900 N)i^+(300 N)j^+(1350 N)k^]

Substitute, (14 m)i^+(2 m)j^+(5 m)k^ for d→BD, (15 m) for dBD, 1500 N for F2 in equation (IV).

T→BD=([(14 m)i^+(2 m)j^+(5 m)k^](15 m))(1500 N)=[(1400 N)i^+(200 N)j^+(500 N)k^]

Substitute, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC in equation (V).

R→=[(1400 N)i^+(200 N)j^+(500 N)k^]+[(900 N)i^+(300 N)j^+(1350 N)k^]=[(2300 N)i^+(500 N)j^+(1850 N)k^]

The magnitude of the resultant force,

|R→|=(2300 N)2+(500 N)2+(1850 N)2=2993.7 N

Thus, the resultant force is (2300N)i^+(500 N)j^+(1850 N)k^_.

Answer 13

(b)

To determine

The pitch of the wrench.

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

Explanation of Solution

Write the equation of pitch of the wrench.

p=M1R (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

Answer 14

(b)

To determine

The pitch of the wrench.

Answer 15

(b)

Expert Solution

Answer to Problem 3.139P

The pitch of the wrench is −0.201 m_.

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 (VI)

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 (VII)

Here, the constant is λaxis.

Write the expression for the constant is,

λaxis=R→R

Write the equation of momentum.

M→oR=[(r→A×T→AC)+(r→B×T→BD)] (VIII)

Here, the momentum is M→oR, the distance is r→A,r→B.

Rewrite the expression for the momentum of the wrench is,

M→1=(R→R)[(r→A×T→AC)+(r→B×T→BD)] (IX)

Conclusion:

Substitute, [(2300N)i^+(500 N)j^+(1850 N)k^] for R→, 2993.7 N for R, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (IX).

M1=([(2300N)i^+(500 N)j^+(1850 N)k^](2993.7 N))[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]=−601.26 N-m

Substitute, −601.26 N-m for M1, (2993.7 N) for R in equation (VI)

p=(−601.26 N-m)(2993.7 N)=−0.201 m

Thus, the pitch of the wrench is −0.201 m_.

Answer 20

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Answer 21

(c)

To determine

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

Answer 22

(c)

Expert Solution

Answer to Problem 3.139P

The axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

EBK <x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>VECTOR</x-custom-btb-me> MECHANICS FOR ENGINEERS: STA, Chapter 3.4, Problem 3.139P

Refer fig 1,

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

M→oR=M→1+M→2 (X)

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→ (XI)

Here, the position vector is r→Q/O.

Write the expression for the position vector is,

r→Q/O=(yj^+zk^)

Here, the coordinates are y,z.

Conclusion:

Substitute, (12 m)k^ for r→A, [(900 N)i^+(300 N)j^+(1350 N)k^] for T→AC, (9 m)i^ for r→B, [(1400 N)i^+(200 N)j^+(500 N)k^] for T→BD in equation (VIII).

MoR=[((12 m)k^×[(900 N)i^+(300 N)j^+(1350 N)k^])+((9 m)i^×[(1400 N)i^+(200 N)j^+(500 N)k^])]=−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]

Substitute, −[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^] for M→oR, [(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^] for M→1 in equation (X).

−[(3600 N-m)i^+(6300 N-m)j^+(1800 N-m)k^]=[(−461.93 N-m)i^+(−100.42 N-m)j^+(−371.56 N-m)k^]+M→2M→2=[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]

Substitute, [(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^] for M→2, [(2300N)i^+(500 N)j^+(1850N)k^] for R→, (xi^+zk^) for r→Q/O in equation (XI).

[(−3138.1 N-m)i^+(6400.4 N-m)j^+(2171.6 N-m)k^]=(yj^+zk^)×[(2300N)i^+(500 N)j^+(1850N)k^]

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

y=−0.944m, z=2.78 m

Therefore, the axis of wrench intersects the yz-plane at y=−0.944m, z=2.78 m_.

Answer 27

Ch. 3.1 - 3.1 A crate of mass 80 kg is held in the position...Ch. 3.1 - 3.2 A crate of mass 80 kg is held in the position...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 - Prob. 3.6P Ch. 3.1 - Prob. 3.7P Ch. 3.1 - Prob. 3.8P 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.

Concept explainers

Videos

The resultant force.

Answer to Problem 3.139P

Explanation of Solution

The pitch of the wrench.

Answer to Problem 3.139P

Explanation of Solution

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

Answer to Problem 3.139P

Explanation of Solution

Want to see more full solutions like this?

Chapter 3 Solutions