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 12.28P

Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.

Fig. P12.28

Chapter 12.1, Problem 12.28P, Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks

(a)

Expert Solution
Check Mark
To determine

Find the magnitude of the force P.

Answer to Problem 12.28P

The magnitude of the force P is 10.00N_.

Explanation of Solution

Given information:

The mass of block A (mA) is 10 kg.

The mass of blocks B (mB) is 5 kg.

The mass of blocks C (mC) is 5 kg.

The initially block B [(yB)0=0] is at rest.

The distance of movement for block B (yB) is 3 m.

The time taken by block B to move 3 m (t) is 2 s.

Calculation:

Sketch the system with position of blocks as shown in Figure 1.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.28P , additional homework tip  1

Write the general equation of weight (W):

W=mg

Here, m is the mass, g is the acceleration due to gravity.

Refer Figure (1).

Consider the position of y be positive downward.

Consider the constraint of cord AD.

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

yA+yD=constant (1)

Here, yA is the length of cord connecting block A and yD is the length of cord connecting block D.

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

vA+vD=constant (2)

Here, vA is the velocity of the block A and vD is the velocity of the block D.

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

aA+aD=0aD=aA (3)

Here, aA is the acceleration of block A and aD acceleration of block D.

Consider the constraint of cord BC.

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

(yByD)+(yCyD)=constantyB+yC2yD=constant (4)

Here,yByD is the length of cord connecting block B and yCyD is the length of cord connecting block C.

Differentiate Equation (4) with respect to t to write velocity of the blocks.

vB+vC2vD=constant (5)

Here, vB is the velocity of the block B and vC is the velocity of the block C.

Differentiate Equation (5) with respect to t to write acceleration of the blocks.

aB+aC2aD=0 (6)

Here, aB is the acceleration of block B and aC acceleration of block C.

Substitute aA for aD.

aB+aC2(aA)=02aA+aB+aC=0 (7)

The motion of blocks is uniform.

Find the acceleration of block B (aB) using general kinematic equation:

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

Here, (vB)0 is the initial velocity of block B.

Substitute 0 for (vB)0.

yB=(yB)0+0+12aBt212aBt2=yB(yB)0aB=2[yB(yB)0]t2

Substitute 3 m for yB, 0 for (yB)0, and 2 s for t.

aB=2[30]22=1.5m/s2

Sketch the free body diagram of pulley D as shown in Figure 2.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.28P , additional homework tip  2

Refer Figure (2),

Consider equilibrium along y-axis.

ΣFy=02TBCTAD=0TAD=2TBC (8)

Here, TBC is the tension in the cord BC and TAD is the tension in the cord AD.

Sketch the free body diagram of block A as shown in Figure (3).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.28P , additional homework tip  3

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

ΣFy=mAaAWATAD=mAaAaA=WATADmA

Here, WA is the weight of block A.

Substitute mAg for WA and 2TBC for TAD.

aA=mAg2TBCmA (9)

Sketch the free body diagram of block C as shown in Figure (4).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.28P , additional homework tip  4

Refer Figure (4),

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

ΣFy=mCaCWCTBC=mCaCaC=WCTBCmC

Here, WC is the weight of block C.

Substitute mCg for WC.

aC=mCgTBCmC (10)

Find the tension in the cord BC using Equation (7).

2aA+aB+aC=0

Substitute mAg2TBCmA for aA, 1.5m/s2 for aB, and mCgTBCmC for aC.

2(mAg2TBCmA)+1.5+mCgTBCmC=0

Substitute 10 kg for mA, 9.81m/s2 for g, and 5 kg for mC.

2(10×9.812TBC10)+1.5+5×9.81TBC5=098.12TBC5+49.05TBC5=1.598.12TBC+49.05TBC=7.5TBC=51.55N

Sketch the free body diagram of block B as shown in Figure (5).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 12.1, Problem 12.28P , additional homework tip  5

Refer Figure (5),

Find the magnitude of the force P.

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

ΣFy=mBaBWB+PTBC=mBaB

Here, WB is the weight of block B.

Substitute mBg for WB.

mBg+PTBC=mBaB

Substitute 5 kg for mB, 9.81m/s2 for g, 51.55 N for P, and 1.5m/s2 for aB.

5×9.81+P51.55=5×1.5P=10.00N

Thus, the magnitude of the force P is 10.00N_.

(b)

Expert Solution
Check Mark
To determine

Find the tension in the cord AD.

Answer to Problem 12.28P

The tension in the cord AD is 103.1N_.

Explanation of Solution

Calculation:

Find the tension in the cord AD using Equation (8).

TAD=2TBC

Substitute 51.55N for TBC.

TAD=2(51.55)=103.1N

Thus, the tension in the cord is 103.1N_.

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