VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Chapter 12.1, Problem 12.15P

(a)

To determine

Find the acceleration of block A for each system.

(a)

Expert Solution
Check Mark

Answer to Problem 12.15P

The acceleration of block A for system 1 is 10.73ft/s2_.

The acceleration of block A for system 2 is 16.10ft/s2_.

The acceleration of block A for system 1 is 0.740ft/s2_.

Explanation of Solution

Calculation:

Sketch the general diagram of systems as shown in Figure (1).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.15P , additional homework tip  1

Write total length of cable connecting block A and block B.

yA+yB=constant (1)

Here,yA is the length of cable connecting block A and yB is the length of cable connecting block B.

Differentiate Equation (1) with respect to t to write velocity of the blocks.

vA+vB=0 (2)

Here, vA is the velocity of the block A and vB is the velocity of the block B.

Differentiate Equation (2) with respect to t to write acceleration of the blocks.

aA+aB=0aB=aA (3)

First of all check the required static friction with static friction to maintain equilibrium.

Sketch the free body diagram and kinetic diagram of block A as shown in Figure (2).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.15P , additional homework tip  2

Refer Figure (2).

Consider downward direction as positive.

Apply Newton’s law of motion along y-axis.

ΣFy=myaAWAT=WAgaAT=WAWAgaA (4)

Here, T is the tension in the cable, WA is the weight of cable A, g is the acceleration due to gravity, and aA acceleration of block A.

Sketch the free body diagram and kinetic diagram of block B as shown in Figure (3).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.15P , additional homework tip  3

Refer Figure (3).

Consider downward direction as positive.

Apply Newton’s law of motion along y-axis.

Find the equation of acceleration of block A.

ΣFy=myaBP+WBT=WBgaB (5)

Here, T is the tension in the cable, WB is the weight of cable B, and aB acceleration of block B.

Substitute WAWAgaA for T and aA for aB

P+WBWA+WAgaA=WBg(aA)Pg+WBgWAg+WAaAg=WBaAgPg+WBgWAg+WAaA=WBaAWAaA+WBaA=(WAWBP)g

aA(WA+WB)=(WAWBP)gaA=(WAWBP)g(WA+WB) (6)

The initial velocity of block A is zero.

Find the equation of velocity of block A using kinematics:

vA2(vA)02=2aA[yA(yA)0]

Here, (vA)0 is the initial velocity of block A and (yA)0 is the length of movement of block A at 0 second.

Substitute At 0 for (vA)0 and 0 for At (yA)0.

vA2(vA)02=2aA[yA(yA)0]vA2=2aAyAvA=2aAyA (7)

Find the equation of time required for block A to reach any velocity.

vA=aAtt=vAaA (8)

Find the acceleration of block A [(aA)1] for system 1 using equation 6:

(aA)1=(WA+WBP)g(WA+WB)

Substitute 200 lb for WA, 100 lb for WB, 0 for P, and 32.2ft/s2 for g.

(aA)1=(2001000)(32.2)(200+100)=10.73ft/s2

Therefore, the acceleration of block A for system 1 is 10.73ft/s2_.

Find the acceleration of block A [(aA)2] for system 2 using equation 6:

(aA)2=(WA+WBP)g(WA+WB)

Substitute 200 lb for WA, 0 for WB, 100 lb for P, and 32.2ft/s2 for g.

(aA)2=(2000100)(32.2)(200+0)=16.10ft/s2

Therefore, the acceleration of block A for system 2 is 16.10ft/s2_.

Find the acceleration of block A [(aA)3] for system 2 using equation 6:

(aA)3=(WA+WBP)g(WA+WB)

Substitute 2200 lb for WA, 2100 for WB, 0 for P, and 32.2ft/s2 for g.

(aA)3=(220021000)(32.2)(2200+2100)=0.749ft/s2

Therefore, the acceleration of block A for system 2 is 0.749ft/s2_.

(b)

To determine

Find the velocity of block A for each system after it has moved through 10 ft

(b)

Expert Solution
Check Mark

Answer to Problem 12.15P

The velocity of block A for system 1 after it has moved through 10 ft is 14.65ft/s_.

The velocity of block A for system 2 after it has moved through 10 ft is 17.94ft/s_.

The velocity of block A for system 3 after it has moved through 10 ft is 3.87ft/s_.

Explanation of Solution

Calculation:

Find the velocity of block A for system 1 [(vA)1] after it has moved through 10 ft using Equation (7):

(vA)1=2(aA)1yA

Substitute 10.73ft/s2 for (aA)1 and 10 ft for yA

(vA)1=2×10.73×10=14.65ft/s

Thus, the velocity of block A for system 1 after it has moved through 10 ft is 14.65ft/s_.

Find the velocity of block A for system 2 [(vA)2] after it has moved through 10 ft using Equation (7):

(vA)2=2(aA)2yA

Substitute 16.10ft/s2 for (aA)1 and 10 ft for yA

(vA)2=2×16.10×10=17.94ft/s2

Thus, the velocity of block A for system 2 after it has moved through 10 ft is 17.94ft/s_.

Find the velocity of block A for system 3 [(vA)3] after it has moved through 10 ft using Equation (7):

(vA)3=2(aA)1yA

Substitute 0.749ft/s2 for (aA)1 and 10 ft for yA

(vA)1=2×0.749×10=3.87ft/s

Thus, the velocity of block A for system 3 after it has moved through 10 ft is 3.87ft/s_.

(c)

To determine

Find the time required for block A to reach a velocity of 20 ft/s

(c)

Expert Solution
Check Mark

Answer to Problem 12.15P

The time required for block A for system 1 to reach a velocity of 20 ft/s is 1.864s_.

The time required for block A for system 2 to reach a velocity of 20 ft/s is 1.242s_.

The time required for block A for system 3 to reach a velocity of 20 ft/s is 26.7s_.

Explanation of Solution

Calculation:

Find the time required for block A for system 1 [(tA)1] to reach a velocity of 20 ft/s using Equation (8):

(tA)1=vA(aA)1

Substitute 20ft/s for vA and 10.73ft/s2 for (aA)1.

(tA)1=2010.73=1.864s

Thus, the time of required for block A for system 1 to reach a velocity of 20 ft/s is 1.864s_..

Find the time required for block A for system 2 [(tA)2] to reach a velocity of 20 ft/s using Equation (8):

(tA)2=vA(aA)2

Substitute 20ft/s for vA and 16.10ft/s2 for (aA)2.

(tA)2=2016.10=1.242s

Thus, the time of required for block A for system 2 to reach a velocity of 20 ft/s is 1.242s_..

Find the time required for block A for system 3 [(tA)3] to reach a velocity of 20 ft/s using Equation (8):

(tA)3=vA(aA)3

Substitute 20ft/s for vA and 0.749ft/s2 for (aA)3.

(tA)1=200.749=26.7s

Thus, the time of required for block A for system 3 to reach a velocity of 20 ft/s is 26.7s_..

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Chapter 12 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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