Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 12.1, Problem 12.18P

Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μ s = 0.20 and μ k = 0.15 . If P = 0 , determine (a) the acceleration of block B, (b) the tension in the cord.

  Chapter 12.1, Problem 12.18P, Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between

Expert Solution
Check Mark
To determine

(a)

Acceleration of block B

Answer to Problem 12.18P

Acceleration of block B:

aB=0.986m/s2

Explanation of Solution

Given:

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  1

mA=40kg,mB=8kg

μs=0.20,μk=0.15

P=0

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  2

From the constraint of the cord:

2xA+xB/A=constant

Then

2vA+vB/A=0

And

2aA+aB/A=0

Now

aB=aA+aB/A

Then

aB=aA+(2aA)

Or:

aB=aA   ... (1)

First, we determine if the blocks will move for the given value of θ. Thus, we seek the value of θ for which the blocks are in impeding motion, with the impeding motion of A down the incline.

For B:

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  3

+Fy=0 :

NABWBcosθ=0

Or:

NAB=WBcosθ

Now

FAB=μsNAB

FAB=0.2mBgcosθ

+Fx=0

T+FAB+WBsinθ=0

Or:

T=mBg(0.2cosθ+WBsinθ)

For A:

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  4

+Fy=0 :

NANABWAcosθ=0

Or:

NA=(mA+mB)gcosθ

Now

FA=μsNAFA=0.2(mA+mB)gcosθ

+Fx=0 :

TFAFAB+WAsinθ=0

T=mAgsinθ0.2(mA+mB)gcosθ0.2mBgcosθ

T=g[mAsinθ0.2(mA+mB)gcosθ]

Equating the two expressions for T,

mBg(0.2cosθ+sinθ)=g[mAsinθ0.2(mA+mB)gcosθ]

Or:

8(0.2+tanθ)=g[40tanθ0.2(40+2×8)]

Or:

tanθ=0.4

Or:

θ=21.8, for impeding motion

Since θ<25, the blocks will move.

Now, consider the motion of the blocks.

For B:

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  5

+Fy=0 :

NABWBcos25=0

Or:

NAB=mBgcos25

Sliding:

FAB=μkNAB=0.15mBcos25

+Fx=mBaB

T+FAB+WBsin25=mBaB

Or:

T=mB[g(0.15cos25+sin25)aB]

T=8[9.81(0.15cos25+sin25)aB]

T=8(5.47952aB)

For A:

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  6

+Fy=0 :

NANABWAcos25=0

Or:

NA=(mA+mB)gcos25

Sliding:

FA=μkNA=0.15(mA+mB)gcos25

+Fx=mAaA :

TFAFAB+WAsin25=mAaA

Substituting and using Eq. (1):

T=mAgsin250.15(mA+mB)gcos250.15mBgcos25mA(aB)

T=g[mAsin250.15(mA+mB)cos25]+mAaB

T=9.81[40sin250.15(40+2×8)cos25]+40aB

T=91.15202+40aB

Equating the two expressions for T,

8(5.47952aB)=91.15202+40aB

Or:

aB=0.98575m/s2

aB=0.986m/s2

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  7

Expert Solution
Check Mark
To determine

(b)

Tension in the cable

Answer to Problem 12.18P

Tension in the cable:

T=51.7N

Explanation of Solution

Given:

Vector Mechanics for Engineers: Dynamics, Chapter 12.1, Problem 12.18P , additional homework tip  8

mA=40kg,mB=8kg

μs=0.20,μk=0.15

P=0

T=8(5.47952aB)

T=8[5.47952(0.98575)]

Or:

T=51.7N

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