The system shown starts from rest, and the length of the upper cord is adjusted so that A , B , and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine ( a ) the accelerations of A and B if a B > 10 mm/s 2 ( b ) the change in position of block D when the velocity of block C is 600 mm/s upward.

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Answer 1

Textbook Question

Chapter 11.2, Problem 11.60P

The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if a B > 10 mm/s² (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

Chapter 11.2, Problem 11.60P, The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C

Expert Solution

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Expert Solution

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

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Answer 2

Textbook Question

Chapter 11.2, Problem 11.60P

The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if a B > 10 mm/s² (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

Chapter 11.2, Problem 11.60P, The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C

Expert Solution

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Expert Solution

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

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Answer 3

Textbook Question

Chapter 11.2, Problem 11.60P

The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if a B > 10 mm/s² (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

Chapter 11.2, Problem 11.60P, The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C

Expert Solution

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Expert Solution

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

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Answer 4

Textbook Question

Chapter 11.2, Problem 11.60P

The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if a B > 10 mm/s² (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

Chapter 11.2, Problem 11.60P, The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C

Expert Solution

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Expert Solution

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Expert Solution

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

Answer 8

Expert Solution

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The acceleration of ‘A’ and ‘B’ if aB>10 mm/s2.

Answer 13

Answer to Problem 11.60P

aA=20 mm/s2aB=40 mm/s2

Answer 14

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 1

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

Calculation:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 2

In above system, the length of each cable is constant

For cable 1,

2yA+2yB+yC=C

Differentiate above equation twice,

2vA+2vB+vC=02aA+2aB+aC=0

For cable 2,

(yD−yA)+(yD−yB)=C2yD−yA−yB=C

Differentiate above equation twice,

2vD−vA−vB=02aD−aA−aB=0

For block ‘A’

yA=yA0+vA0t+12aAt2yA=yA0+0+12aAt2

For block ‘C’

yC=yC0+vC0t+12aCt2yC=yC0+0+12aCt2

Therefore,

yC/A=yC−yA=12t2(aC−aA)

At, t=2 s

−280 mm=12(2 s)2(aC−aA) aC=aA−140

But we know that,

2aA+2aB+aC=0

Then,

2aA+2aB+aC=02aA+2aB+(aA−140)=0 aA=(140−2aB)3

For block ‘A’

yA=yA0+vA0t+12aAt2yA−yA0=0+12aAt2160 mm=12aAt2

For block ‘B’

yB=yB0+vB0t+12aBt2yB−yB0=0+12aBt2320 mm=12aBt2

Then,

160 mm=12(aB−aA)t2

But we have already found,

80 mm=(aB−aA)t

Therefore,

t=4 s

Then, for block ‘A’

160 mm=12aA(4 s)2aA=20 mm/s2

Then, for collar ‘B’

320 mm=12aB(4 s)2aB=40 mm/s2

Conclusion:

The acceleration of ‘A’ and ‘B’ are

aA=20 mm/s2aB=40 mm/s2

Answer 15

Expert Solution

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is 600 mm/s

Answer 20

Answer to Problem 11.60P

yD−yD0=375 mm Downwards

Answer 21

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 11.2, Problem 11.60P , additional homework tip 3

At t=0

yA0=yB0=yC0

Each component moves in constant acceleration

After t=2 s

The relative change in position of block ‘C’ with respect to block ‘A’ is yC/A=280 mm in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is 80 mm/s in downward direction.

The displacement of block ‘A’ is 160 mm downwards.

The displacement of collar ‘B is 320 mm downwards.

For a uniformly accelerated motion,

x=x0+v0t+12at2

In above equation,

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

According to sub part a,

We have found

aC=aA−140

2aD−aA−aB=0

Bu we know that,

aA=20 mm/s

Therefore,

aC=−120 mm/s2

2aD−aA−aB=02aD−20 mm/s2−40 mm/s2=0 aD=30 mm/s2

When,

vC=−600 mm/s

Therefore,

vC=vC0+aCt−600 mm/s=0+(−120 mm/s2)t t=5 s

To find the change in distance of block ‘D’

yD=yD0+vD0t+12aDt2yD−yD0=0+12(30 mm/s2)(5 s)2yD−yD0=375 mm

Conclusion:

The change in distance of block ‘D’ is equal to yD−yD0=375 mm in downward direction

Answer 22

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Answer to Problem 11.60P

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