Find whether the slipping occurs between the belt and either cylinder.

Question

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Answer 1

Question

Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

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Answer 2

Question

Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

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

Answer 3

Question

Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

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 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

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Answer 5

Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Answer 6

Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

Answer 7

Chapter 16.1, Problem 16.40P

(a)

To determine

Find whether the slipping occurs between the belt and either cylinder.

Answer 8

(a)

Expert Solution

Explanation of Solution

The force pulled between cylinders A and B (P) is $2.00 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Consider the acceleration due to gravity (g) as $32.2 ft/s2$ .

Convert the unit of the radius of the cylinder A $(rA)$ :

$rA=4 in.×1 ft12 in.=13 ft$

Convert the unit of the radius of the cylinder B $(rB)$ :

$rB=8 in.×1 ft12 in.=23 ft$

Consider that no slipping occurs.

Calculate the acceleration of the belt $(abelt)$ :

$abelt=rAαA=rBαBabelt=αA3=2αB3$

$αA3=2αB3αA=2αBαB=αA2$

Calculate the mass of the cylinder A $(mA)$ :

$mA=WAg$

Substitute $5 lb$ for $WA$ and $32.2 ft/s2$ for g.

$mA=532.2 lb⋅s2/ft$

Calculate the mass of the cylinder B $(mB)$ :

$mB=WBg$

Substitute $20 lb$ for $WB$ and $32.2 ft/s2$ for g.

$mB=2032.2 lb⋅s2/ft$

Calculate the mass moment of inertia of the cylinder A $(I¯A)$ :

$I¯A=12mArA2$

Substitute $532.2 lb⋅s2/ft$ for $mA$ and $13 ft$ for $rA$ .

$I¯A=12×532.2×(13)2=8.6266×10−3 lb⋅s2⋅ft$

Calculate the mass moment of inertia of the cylinder B $(I¯B)$ :

$I¯B=12mBrB2$

Substitute $2032.2 lb⋅s2/ft$ for $mB$ and $23 ft$ for $rB$ .

$I¯B=12×2032.2×(23)2=138.0262×10−3 lb⋅s2⋅ft$

Show the free body diagram of the cylinder A as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 1

Here, $FA$ is the horizontal force of the cylinder A and $αA$ is the angular acceleration of the cylinder A.

Refer to Figure 1.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FArA)=I¯AαA$

Substitute $13 ft$ for $rA$ and $8.6266×10−3 lb⋅s2⋅ft$ for $I¯A$ .

$(FA×13)=8.6266×10−3αA13FA=8.6266×10−3αAFA=8.6266×10−3αA×3FA=25.8798×10−3αA$ (1)

Show the free body diagram of the cylinder B as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 2

Here, $FB$ is the horizontal force of the cylinder B and $αB$ is the angular acceleration of the cylinder B.

Refer to Figure 2.

Calculate the moment about point G by applying the equation of equilibrium:

$∑MG=IGα(FBrB)=I¯BαB$

Substitute $23 ft$ for $rB$ , $αA2$ for $αB$ , and $138.0262×10−3 lb⋅s2⋅ft$ for $I¯B$ .

$(FB×23)=138.0262×10−3×αA223FB=69.0131×10−3αAFB=69.0131×10−3αA×32FB=103.51965×10−3αA$ (2)

Show the free body diagram of the belt as in Figure 3.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.40P , additional homework tip 3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$∑Fx=02.00−FA−FB=0FA+FB=2.00$ (3)

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $25.8798×10−3αA$ for $FA$ and $103.51965×10−3αA$ for $FB$ in Equation (3)

$25.8798×10−3αA+103.51965×10−3αA=2.00129.4125×10−3αA=2.00αA=2.00129.4125×10−3αA=15.46 rad/s2$

Calculate the horizontal force of the cylinder A $(FA)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (1).

$FA=25.8798×10−3×15.46=0.4 lb$

Calculate the horizontal force of the cylinder B $(FB)$ :

Substitute $15.46 rad/s2$ for $αA$ in Equation (2).

$FB=103.51965×10−3×15.46=1.6 lb$

Calculate the magnitude of the friction force $(Fm)$ :

$Fm=μsWA$

Substitute $5 lb$ for $WA$ and $0.50$ for $μs$ .

$Fm=0.50×5=2.5 lb$

The horizontal forces of the cylinder A and B are greater than the magnitude of the friction force $(FA and FB<Fm)$ .

Therefore, there is no slipping occurs between cylinders and belt.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Answer 13

(b)

To determine

Find the angular acceleration of each cylinder $(αA and αB)$ .

Answer 14

(b)

Expert Solution

Answer to Problem 16.40P

The angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

The force pulled between cylinders A and B (P) is $3.6 lb$ .

The weight of the cylinder A $(WA)$ is $5 lb$ .

The weight of the cylinder B $(WB)$ is $20 lb$ .

The coefficient of the static friction $(μs)$ is $0.50$ .

The coefficient of the kinetic friction $(μk)$ is $0.40$ .

The radius of the cylinder A $(rA)$ is $4 in.$ .

The radius of the cylinder B $(rB)$ is $8 in.$ .

Calculation:

Refer the part (a).

Consider the no slipping occur at cylinder B.

Therefore, the angular acceleration of the cylinder B is $αB=12αA$

Calculate the angular acceleration of the cylinder A $(αA)$ :

Substitute $1.6 lb$ for $FB$ in Equation (2).

$1.6=103.51965×10−3αAαA=1.6103.51965×10−3αA=15.46 rad/s2$

Calculate the angular acceleration of the cylinder B $(αB)$ :

$αB=12αA$

Substitute $15.46 rad/s2$ for $αA$ .

$αB=12×15.46=7.73 rad/s2$

Hence, the angular acceleration of each cylinder $(αA and αB)$ are $15.46 rad/s2_$ and $7.73 rad/s2_$ .

Answer 19

Ch. 16.1 - Two pendulums, A and B, with the masses and...Ch. 16.1 - Two pendulums, A and B, with the masses and...Ch. 16.1 - Two solid cylinders, A and B, have the same mass m...Ch. 16.1 - Prob. 16.1FBP Ch. 16.1 - Prob. 16.2FBP Ch. 16.1 - Prob. 16.3FBP Ch. 16.1 - The 400-lb crate shown is lowered by means of two...Ch. 16.1 - A 60-lb uniform thin panel is placed in a truck...Ch. 16.1 - A 60-lb uniform thin panel is placed in a truck...Ch. 16.1 - 16.3 Knowing that the coefficient of static...

Find whether the slipping occurs between the belt and either cylinder.

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Find whether the slipping occurs between the belt and either cylinder.

Explanation of Solution

Find the angular acceleration of each cylinder $(αA and αB)$ .

Answer to Problem 16.40P

Explanation of Solution

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