The perpendicular distance ( d ) between rod AB and the line of action of P .

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Answer 1

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

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Homework#8

Sign in PDF Lecture W09.pdf PDF MMB241 - Tutorial L9.pdf File C:/Users/KHULEKANI/Desktop/mmb241/MMB241%20-%20Tutorial%20L9.pdf II! Draw | I│Alla | Ask Copilot + of 4 D Topic: Kinetics of Particles: - Forces in dynamic system, Free body diagram, newton's laws of motion, and equations of motion. TQ1. The 10-kg block is subjected to the forces shown. In each case, determine its velocity when t=2s if v 0 when t=0 500 N F = (201) N 300 N (b) TQ2. The 10-kg block is subjected to the forces shown. In each case, determine its velocity at s-8 m if v = 3 m/s at s=0. Motion occurs to the right. 40 N F = (2.5 s) N 200 N 30 N (b) TQ3. Determine the initial acceleration of the 10-kg smooth collar. The spring has an unstretched length of 1 m. 1 σ Q ☆ Q 6 ا الى ☑

Sign in PDF Lecture W09.pdf PDF MMB241 - Tutorial L9.pdf File C:/Users/KHULEKANI/Desktop/mmb241/MMB241%20-%20Tutorial%20L9.pdf II! Draw | I│Alla | Ask Copilot + 4 of 4 | D TQ9. If motor M exerts a force of F (10t 2 + 100) N determine the velocity of the 25-kg crate when t kinetic friction between the crate and the plane are μs The crate is initially at rest. on the cable, where t is in seconds, 4s. The coefficients of static and 0.3 and μk = 0.25, respectively. M 3 TQ10. The spring has a stiffness k = 200 N/m and is unstretched when the 25-kg block is at A. Determine the acceleration of the block when s = 0.4 m. The contact surface between the block and the plane is smooth. 0.3 m F= 100 N F= 100 N k = 200 N/m σ Q Q ☆ ا الى 6 ☑

Answer 2

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

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Answer 3

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

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Answer 4

Question

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

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Answer 5

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Answer 6

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Answer 7

Chapter 3.2, Problem 3.64P

To determine

The perpendicular distance (d) between rod AB and the line of action of P.

Answer 8

Expert Solution & Answer

Answer to Problem 3.64P

The perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The length of the vertical rod CD $(lCD)$ is 23 inch.

The length of the rod AB $(lAB)$ is 50 inch.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 inch.

The distance between the points O to A (y) is 24 inch.

The vertical height of OA $(hOA)$ is 30 inch.

The vertical height of OH $(hOH)$ is 17 inch.

The horizontal distance between point O and midpoint C $(dOC)$ is 16 inch.

The horizontal distance between midpoint C and point G over x axis $(dCG)$ is 21 inch.

The horizontal distance between point B and midpoint C $(dBC)$ is 12 inch.

The horizontal distance between midpoint C and point G over z axis (h) is 18 inch.

Calculation:

Calculate the position vector of AB using the relation:

$AB⇀=xi−hOAj−yk$

Substitute 32 in. for x, 30 in. for $hOA$ , and 24 in. for y.

$AB⇀=32i−30j−24kAB=(32)2+(−30)2+(−24)2AB=50 in.$

Calculate the unit vector of AB $(λAB)$ using the formula:

$λAB=AB⇀AB$

Substitute $32i−30j−24k$ for $AB⇀$ and 50 in. for AB.

$λAB=32i−30j−24k50=0.64i−0.6j−0.48k$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $(rG/B)$ using the relation:

$rG/B=[x−(dOC+dCG)]i+(dBC+h)k$

Substitute 32 in. for x, 16 in. for $dOC$ , 21 in. for $dCG$ , 12 in. for $dBC$ , and 18 in. for h.

$rG/B=[32−(16+21)]i+(12+18)k=5i+30k$

Calculate the position vector of DG using the relation:

$DG⇀=dCGi−(hOA2+lCD)j+hk$

Substitute 21 in. for $dCG$ , 30 in. for $hOA$ , 23 in. for $lCD$ , and 18 in. for h.

$DG⇀=21i−(302+23)j+18kDG⇀=21i−38j+18kDG=(21)2+(−38)2+(18)2DG=47 in.$

Calculate the force $(P)$ using the formula:

$P=PDG⇀DG$

Substitute 235 lb for P, $21i−38j+18k$ for $DG⇀$ , and 47 in. for DG.

$P=235×21i−38j+18k47=235×(0.447i−0.808j+0.383k)=105i−189.9j+90k≈105i−190j+90k$

Calculate the position vector at P $(λP)$ using the relation:

$λP=PP$

Substitute $105i−190j+90k$ for P and 235 lb for P.

$λP=105i−190j+90k235=0.447i−0.808j+0.383k$

Calculate the angle between AB and P $(θ)$ using the formula:

$cosθ=λAB⋅λP$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ and $0.447i−0.808j+0.383k$ for $λP$ .

$cosθ=(0.64i−0.6j−0.48k)⋅(0.447i−0.808j+0.383k)cosθ=0.28608+0.4848−0.18384cosθ=0.5870θ=cos−1(0.5870)θ=54.04°$

Calculate the moment about AB $(MAB)$ using the relation:

$MAB=λAB⋅(rG/B×P)$

Substitute $0.64i−0.6j−0.48k$ for $λAB$ , $5i+30k$ for $rG/B$ and $105i−190j+90k$ for P.

$MAB=|0.64−0.6−0.485030105−19090|=[0.64(0−(−5700))]−[(−0.6)(450−3150)]−[0.48(−950−0)]=3648−1620+456=2484 lb⋅in.×112ftin.=207 lb⋅ft$

Calculate the perpendicular distance (d) between rod AB and the line of action of P:

Take moment about AB:

$MAB=(Psinθ)d$

Substitute 235 lb for P and $54.04°$ for $θ$ , and $207lb⋅ft$ for $MAB$ .

$207=(235sin54.04)dd=207235sin54.04d=1.088 ft×12in.ftd=13.06 in.$

Thus, the perpendicular distance (d) between rod AB and the line of action of P is $13.06 in._$ .

Answer 12

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...

The perpendicular distance ( d ) between rod AB and the line of action of P .

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The perpendicular distance (d) between rod AB and the line of action of P.

Answer to Problem 3.64P

Explanation of Solution

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