The plate shown moves in the xy plane. Knowing that ( v A ) x = 250 mm/s, ( v B ) y = − 450 mm/s and ( v c ) x = − 500 mm/s, determine ( a ) the angular velocity of the plate, ( b ) the velocity of point A .

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Answer 1

Textbook Question

Chapter 15.2, Problem 15.46P

The plate shown moves in the xy plane. Knowing that $( v A ) x = 250$ mm/s, $( v B ) y = − 450$ mm/s and $( v c ) x = − 500$ mm/s, determine (a) the angular velocity of the plate, (b) the velocity of point A.

Chapter 15.2, Problem 15.46P, The plate shown moves in the xy plane. Knowing that (vA)x=250 mm/s, (vB)y=450 mm/s and (vc)x=500

Expert Solution

To determine

(a)

The angular velocity of the plate.

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Expert Solution

To determine

(b)

The velocity of point A.

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

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Answer 2

Textbook Question

Chapter 15.2, Problem 15.46P

The plate shown moves in the xy plane. Knowing that $( v A ) x = 250$ mm/s, $( v B ) y = − 450$ mm/s and $( v c ) x = − 500$ mm/s, determine (a) the angular velocity of the plate, (b) the velocity of point A.

Chapter 15.2, Problem 15.46P, The plate shown moves in the xy plane. Knowing that (vA)x=250 mm/s, (vB)y=450 mm/s and (vc)x=500

Expert Solution

To determine

(a)

The angular velocity of the plate.

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Expert Solution

To determine

(b)

The velocity of point A.

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

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Answer 3

Textbook Question

Chapter 15.2, Problem 15.46P

The plate shown moves in the xy plane. Knowing that $( v A ) x = 250$ mm/s, $( v B ) y = − 450$ mm/s and $( v c ) x = − 500$ mm/s, determine (a) the angular velocity of the plate, (b) the velocity of point A.

Chapter 15.2, Problem 15.46P, The plate shown moves in the xy plane. Knowing that (vA)x=250 mm/s, (vB)y=450 mm/s and (vc)x=500

Expert Solution

To determine

(a)

The angular velocity of the plate.

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Expert Solution

To determine

(b)

The velocity of point A.

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

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Answer 4

Textbook Question

Chapter 15.2, Problem 15.46P

The plate shown moves in the xy plane. Knowing that $( v A ) x = 250$ mm/s, $( v B ) y = − 450$ mm/s and $( v c ) x = − 500$ mm/s, determine (a) the angular velocity of the plate, (b) the velocity of point A.

Chapter 15.2, Problem 15.46P, The plate shown moves in the xy plane. Knowing that (vA)x=250 mm/s, (vB)y=450 mm/s and (vc)x=500

Expert Solution

To determine

(a)

The angular velocity of the plate.

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Expert Solution

To determine

(b)

The velocity of point A.

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The angular velocity of the plate.

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Expert Solution

To determine

(b)

The velocity of point A.

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

Answer 8

Expert Solution

To determine

(a)

The angular velocity of the plate.

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The angular velocity of the plate.

Answer 13

Answer to Problem 15.46P

The angular velocity of the plate is $−5 rad/s$ .

Answer 14

Explanation of Solution

Given Information:

The figure below shows a plate moving in x-y plane.

Vector Mechanics for Engineers: Dynamics, Chapter 15.2, Problem 15.46P

Figure-(1)

The velocity along x- direction of the point A is $250 mm/s$ , the velocity of point B along y -direction is $−450 mm/s$ and the velocity of point C along x -direction is $−500 mm/s$ .

Write the expression for position vector of AB.

$rB/A=(Bx−Ax)i+(By−Ay)j$ ...... (I)

Here, the coordinates of point A is $(Ax, Ay)$ and the coordinates of point B is $(Bx, By)$ .

Write the expression for the velocity at point A.

$vA=(vA)xi+(vA)yj$ ...... (II)

Here, the velocity of point A along x -direction is $(vA)x$ , the velocity of point A along y -direction is $(vA)y$ .

Write the expression for the velocity at point B.

$vB=(vB)xi+(vB)yj$ ...... (III)

Here, the velocity of point B along x -direction is $(vB)x$ , the velocity of point B along y -direction is $(vB)y$ .

Write the expression for the velocity at point C.

$vC=(vC)xi+(vC)yj$ ...... (IV)

Here, the velocity of point C along x -direction is $(vC)x$ , the velocity of point C along y -direction is $(vC)y$ .

Write the expression to calculate the velocity of point B.

$vB=vA+ωk(rB/A)$ ..... (V)

Here, the angular velocity of the plate is $ω$ .

Write the expression for velocity at point C.

$vC=(vA)x+ωk(rC/A)$ ...... (VI)

Write the expression for position vector of CA.

$rC/A=(Cx−Ax)i+(Cy−Ay)j$ ...... (VII)

Here, the coordinates of point C is $(Cx, Cy)$ .

Calculation:

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $150 mm$ for $Bx$ and $150 mm$ for $By$ in Equation (I).

$rB/A=(150 mm−0 mm)i+(150 mm−150 mm)j=(150 mm)i$

Substitute $0 mm$ for $Ax$ , $150 mm$ for $Ay$ , $200 mm$ for $Cx$ and $0 mm$ for $Cy$ in Equation (VII).

$rC/A=(200 mm−0 mm)i+(0 mm−150 mm)j=(200 mm)i−(150 mm)j$

Substitute $(vC)xi+(vC)yj$ for $(vC)$ , $(vA)xi+(vA)yj$ for $vA$ and $(200 mm)i−(150 mm)j$ for $rC/A$ in Equation (VI).

$(vC)xi+(vC)yj=(vA)xi+(vA)yj+ωk((200 mm)i−(150 mm)j)$ ...... (VIII)

Substitute $−500 mm/s$ for $(vC)x$ and $250 mm/s$ for $(vA)x$ in Equation (VIII).

$(−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+ωk((200 mm)i−(150 mm)j)](−500 mm/s)i+(vC)yj=[(250 mm/s)i+(vA)yj+((200ω mm)j−(150ω mm)i)]$ ...... (IX)

Equate the coefficients of $i$ in Equation (IX).

$(−500 mm/s)=(250 mm/s)+(150ω mm)ω=−750 mm/s150 mmω=−5 rad/s$

Conclusion:

The angular velocity of the plate is $−5 rad/s$ and the negative sign indicates the clockwise direction of the angular velocity.

Answer 15

Expert Solution

To determine

(b)

The velocity of point A.

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

The velocity of point A.

Answer 20

Answer to Problem 15.46P

The velocity of point A is $390.51 mm/s$ .

Answer 21

Explanation of Solution

Write the expression for velocity of point A.

$vA=(vA)x2+(vA)y2$ ...... (X)

Calculation:

Substitute $(vB)xi+(vB)yj$ for $vB$

$(vA)xi+(vA)yj$ for $vA$ and $(150 mm)i$ for $rB/A$ in Equation (V).

$(vB)xi+(vB)yj= [(vA)xi+(vA)yj+ωk(150 mm)i](vB)xi+(vB)yj= [(vA)xi+(vA)yj+ω(150 mm)j]$ ...... (XI)

Substitute $−450 mm/s$ for $(vB)y$

$250 mm/s$ for $(vA)x$ and $−5 rad/s$ for $ω$ in Equation (XI).

$(vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(150(−5 rad/s) mm)j](vB)xi+(−450 mm/s)j=[(250 mm/s)i+(vA)yj+(−750 mm/s)j]$ ...... (XII)

Equate the coefficients of j in Equation (XII).

$(−450 mm/s)=(vA)y−750 mm/s(vA)y=300 mm/s$

Substitute $300 mm/s$ for $(vA)y$ and $250 mm/s$ for $(vA)x$ in Equation (X).

$va=(250 mm/s)2+(300 mm/s)2=152500 mm2/s2=390.51 mm/s$

Conclusion:

The velocity of point A is $390.51 mm/s$ .

Answer 22

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The plate shown moves in the xy plane. Knowing that ( v A ) x = 250 mm/s, ( v B ) y = − 450 mm/s and ( v c ) x = − 500 mm/s, determine ( a ) the angular velocity of the plate, ( b ) the velocity of point A .

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Answer to Problem 15.46P

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Answer to Problem 15.46P

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