A 300-N force P is applied at point A of the bell crank shown. ( a ) Compute the moment of the force P about O by resolving it into horizontal and vertical components. ( b ) Using the result of part a , determine the perpendicular distance from O to the line of action of P.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 3, Problem 3.147RP

A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part a, determine the perpendicular distance from O to the line of action of P.

Chapter 3, Problem 3.147RP, A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P

(a)

Expert Solution

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

(b)

Expert Solution

To determine

The perpendicular distance from the point O to $P$ .

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

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

please show a drawing/image and explain how to properly do the question. thanks

For the four-bar- linkage shown in the following figure. BC=68mm, CD=100mm, AD=120mm. Determine the range of AB to make it a crank-rocker mechanism. B D

all of those 4 fi 1)Draw kinematic diagram: 2)DOF: 3)type/name of mechanism 4)evolution:

Answer 2

Textbook Question

Chapter 3, Problem 3.147RP

A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part a, determine the perpendicular distance from O to the line of action of P.

Chapter 3, Problem 3.147RP, A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P

(a)

Expert Solution

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

(b)

Expert Solution

To determine

The perpendicular distance from the point O to $P$ .

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

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

Textbook Question

Chapter 3, Problem 3.147RP

A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part a, determine the perpendicular distance from O to the line of action of P.

Chapter 3, Problem 3.147RP, A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P

(a)

Expert Solution

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

(b)

Expert Solution

To determine

The perpendicular distance from the point O to $P$ .

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

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

Textbook Question

Chapter 3, Problem 3.147RP

A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part a, determine the perpendicular distance from O to the line of action of P.

Chapter 3, Problem 3.147RP, A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P

(a)

Expert Solution

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

(b)

Expert Solution

To determine

The perpendicular distance from the point O to $P$ .

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

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

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

(b)

Expert Solution

To determine

The perpendicular distance from the point O to $P$ .

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Answer 8

(a)

Expert Solution

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The moment of the force P about O

Answer 13

Answer to Problem 3.147RP

The magnitude of the moment is about O is $20.5 N⋅m_$ .

Answer 14

Explanation of Solution

The magnitude of P is $300 N$ .

Write the expression for the vector from O to A.

$rOA=200 mm(1 m1000 mm)cos40° i^+200 mm(1 m1000 mm)sin40° j^=(0.153209 i^+0.128558 j^) m$ (I)

Here $rOA$ is the vector from O to A.

Write the expression for the force vector $F$ .

$F=300 N(sin30°) i^+300 N(cos30°) j^=(150 N) i^+(259.81 N) j^$ (II)

Write the expression for the moment at O.

$MO=(rOA×F)$

Here $MO$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $MO$ .

$MO=(0.153209 i^+0.128558 j^) m×(150 N i^+259.81 N j^)=(39.805 k^−19.2837 k^) N⋅m=(20.521 N⋅m) k^=20.5 N⋅m$

Thus, the magnitude of moment at O is $20.5 N⋅m_$ .

Answer 15

(b)

Expert Solution

To determine

The perpendicular distance from the point O to $P$ .

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The perpendicular distance from the point O to $P$ .

Answer 20

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Answer 21

Explanation of Solution

The magnitude of moment at O is $20.5 N⋅m$ .

Write the expression to calculate the perpendicular distance from the point O to force $P$ .

$Fd=MO$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $MO$ is the magnitude of the moment.

Conclusion:

Substitute $300 N$ for P and $20.5 N⋅m$ for $MO$ in the above equation to calculate d.

$(300 N)d=20.5 N⋅md=20.5 N⋅m300 N=0.0683 m(1000 mm1 m)=68.3 mm$

Thus, the perpendicular distance from the point O to $P$ is $68.3 mm_$ .

Answer 22

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 - 3.6 A 20-lb force is applied to the control rod AB...Ch. 3.1 - 3.7 A 20-lb force is applied to the control rod AB...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...

A 300-N force P is applied at point A of the bell crank shown. ( a ) Compute the moment of the force P about O by resolving it into horizontal and vertical components. ( b ) Using the result of part a , determine the perpendicular distance from O to the line of action of P.

Videos

Answer to Problem 3.147RP

Explanation of Solution

Answer to Problem 3.147RP

Explanation of Solution

Want to see more full solutions like this?

Chapter 3 Solutions