Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s 2 . Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine ( a ) the accelerations of A and B , ( b ) the initial velocities of A and C , ( c ) the change in position of slider block C after 3 s.

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Answer 1

Textbook Question

Chapter 11.2, Problem 11.57P

Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s². Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

Chapter 11.2, Problem 11.57P, Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to

Expert Solution

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Expert Solution

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Expert Solution

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer to Problem 11.57P

xC−xC0=727.5 mm

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

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Answer 2

Textbook Question

Chapter 11.2, Problem 11.57P

Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s². Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

Chapter 11.2, Problem 11.57P, Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to

Expert Solution

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Expert Solution

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Expert Solution

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer to Problem 11.57P

xC−xC0=727.5 mm

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

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Answer 3

Textbook Question

Chapter 11.2, Problem 11.57P

Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s². Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

Chapter 11.2, Problem 11.57P, Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to

Expert Solution

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Expert Solution

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Expert Solution

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer to Problem 11.57P

xC−xC0=727.5 mm

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

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Answer 4

Textbook Question

Chapter 11.2, Problem 11.57P

Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s². Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

Chapter 11.2, Problem 11.57P, Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to

Expert Solution

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Expert Solution

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Expert Solution

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer to Problem 11.57P

xC−xC0=727.5 mm

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Expert Solution

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Expert Solution

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer to Problem 11.57P

xC−xC0=727.5 mm

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

Answer 8

Expert Solution

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The acceleration A and B.

Answer 13

Answer to Problem 11.57P

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Answer 14

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 1

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 2

In above system the total length of rope is constant, therefore,

3yA+4yB+xC=C

Where,

C - Constant

Differentiate above equation,

3vA+4vB+vC=0

Differentiate above equation,

3aA+4aB+aC=0

For block ‘B’

At t=2 s

vB=(vB)0+aBt

Substitute,

480 mm/s=0+aB(2 s) aB=240 mm/s2

Therefore,

3aA+4aB+aC=03aA+4(240 mm/s2)+(75 mm/s2)=0 aA=−345 mm/s2

Conclusion:

Acceleration of block A and B are,

aB=240 mm/s2 Downwards

aA=345 mm/s2 Upwards

Answer 15

Expert Solution

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Initial velocities of A and C.

Answer 20

Answer to Problem 11.57P

vA0=43.33 mm/svC0=130 mm/s

Answer 21

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 3

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

For a uniformly accelerated motion,

v=v0+at

In above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

t - Elapsed time

Calculation:

For block ‘C’

At t=2 s

vC=(vC)0+aCt280 mm/s=(vC)0+(75 mm/s2)(2 s) (vC)0=130 mm/s

According to sub-part a,

We know that

3vA+4vB+vC=0

Therefore,

3vA+4vB+vC=03vA+0+130 mm/s=0 vA0=43.33 mm/s

Conclusion:

Initial velocitiesof block A and C are,

vA0=43.33 mm/svC0=130 mm/s

Answer 22

Expert Solution

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer to Problem 11.57P

xC−xC0=727.5 mm

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

(c)

The change in position of block ‘C’ after 3 s.

Answer 27

Answer to Problem 11.57P

xC−xC0=727.5 mm

Answer 28

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.2, Problem 11.57P , additional homework tip 4

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of 75 mm/s2

After, t=2 s

vB=480 mm/svC=280 mm/s

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 b,

vC0=130 mm/s

For block C,

After 3 s

xC=xC0+vC0t+12aCt2xC−xC0=(130 mm/s)(3 s)+12(75 mm/s2)(3 s)2xC−xC0=727.5 mm

Conclusion:

Change in distance of block C after 3 s is equal to xC−xC0=727.5 mm

Answer 29

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Answer to Problem 11.57P

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Answer to Problem 11.57P

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Answer to Problem 11.57P

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