Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 12.1, Problem 12.28P
To determine

(a)

The magnitude of force P.

Expert Solution
Check Mark

Answer to Problem 12.28P

P=10Ν

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  1

Mass of block A =10kg

Mass of block B=C =5kg

Time t=2sec

Let the position co-ordinate be Y and it is positive towards downwards.

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  2

Now, constraint of cord AD: yA+yD= constant

Then, vA+vD=0vD=vA__________(1) or,

aA+aD=0aD=aA__________(2)

Similarly, constraint of cord BC: (yByD)+(yCyD)= constant

yB+yC2yD= Constant

Then, vB+vC2vD=0 or,

aB+aC2aD=0

Now, from equation (1) and (2);

vB+vC2(vA)=02vA+vB+vC=0 And,

aB+aC2(aA)=02aA+aB+aC=0___________(3)

As all of the forces are constant then, we have uniformly accelerated motions.

yB=(yB)0+(vB)0t+12aBt2 or,

(vB)0=0

Then,

aB=2[yB(yB)0]t2aB=2[3](2)2=1.5m/s2

Now, calculate all the forces for pulley D,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  3

Fy=0

TBC+TBCTAD=02TBCTAD=0or,2TBC=TAD

Now, calculate all the forces for block A, for this the draw the free body and kinetic diagram of the block A:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  4

Fy=may

WATAD=mAaAaA=WATADmAor,aA=WA2TBCmA________(4)

Calculate all the forces for block C, for this the draw the free body and kinetic diagram of block C:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  5

Fy=may

WCTBC=mCaCaC=WCTBCmA________(5)

Calculate all the forces for block B, for this the draw the free body and kinetic diagram of block B:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  6

Fy=may

P+WBTBC=mBaBP=mBaB+TBCWB______(6)

Substituting the values of equation (4) and (5) into equation (3);

2(WA2TBCmA)+aB+(WCTBCmC)=02(mAg2TBCmA)+aB+(mCgTBCmC)=02g4TBCmA+aB+gTBCmC=03g+aB=(4mA+1mC)TBC3(9.81)+1.5=(410+15)TBCTBC(35)=30.93TBC=51.55Ν

Putting all the values in equation (6);

P=mBaB+TBCWBP=5(1.5)+51.555(9.81)P=7.5+51.5549.05P=10Ν

To determine

(b)

The tension in cord AD.

Expert Solution
Check Mark

Answer to Problem 12.28P

TAD=103.1Ν

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  7

Mass of block A =10kg

Mass of block B=C =5kg

Time t=2sec

Let the position co-ordinate be Y and it is positive towards downwards.

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  8

Now, constraint of cord AD: yA+yD= constant

Then, vA+vD=0vD=vA__________(1) or,

aA+aD=0aD=aA__________(2)

Similarly, constraint of cord BC: (yByD)+(yCyD)= constant

yB+yC2yD= Constant

Then, vB+vC2vD=0 or,

aB+aC2aD=0

Now, from equation (1) and (2);

vB+vC2(vA)=02vA+vB+vC=0 And,

aB+aC2(aA)=02aA+aB+aC=0___________(3)

As all of the forces are constant then we have uniformly accelerated motions.

yB=(yB)0+(vB)0t+12aBt2 or,

(vB)0=0

Then,

aB=2[yB(yB)0]t2aB=2[3](2)2=1.5m/s2

Now, calculate all the forces for pulley D,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  9

Fy=0

TBC+TBCTAD=02TBCTAD=0or,2TBC=TAD

Now, calculate all the forces for block A, for this the draw the free body and kinetic diagram of the block A:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  10

Fy=may

WATAD=mAaAaA=WATADmAor,aA=WA2TBCmA________(4)

Calculate all the forces for block C, for this the draw the free body and kinetic diagram of the block C:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.28P , additional homework tip  11

Fy=may

WCTBC=mCaCaC=WCTBCmA________(5)

Substituting the values of equation (4) and (5) into equation (3);

2(WA2TBCmA)+aB+(WCTBCmC)=02(mAg2TBCmA)+aB+(mCgTBCmC)=02g4TBCmA+aB+gTBCmC=03g+aB=(4mA+1mC)TBC3(9.81)+1.5=(410+15)TBCTBC(35)=30.93TBC=51.55Ν

Now, tension in cord AD; TAD=2TBC

TAD=2(51.55)TAD=103.1Ν

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Chapter 12 Solutions

Vector Mechanics For Engineers

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