Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 12.1, Problem 12.19P

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 μs = 0.20 and μk = 0.15. If P = 0, determine (a) the acceleration of block B, (b) the tension in the cord.

Fig. P12.18 and P12.19

Chapter 12.1, Problem 12.19P, Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between

(a)

Expert Solution
Check Mark
To determine

Find the acceleration of block B.

Answer to Problem 12.19P

The acceleration of block B is 0.986m/s2_.

Explanation of Solution

Given information:

The mass of block A (mA) is 40 kg.

The mass of block B (mB) is 8 kg.

The coefficient of static friction between all surfaces of contact (μs) is 0.20.

The coefficient of kinetic friction between all surfaces of contact (μk) is 0.15.

The horizontal load (P) is 40 N.

Calculation:

Write the general equation of weight  (W):

W=mg

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

Consider the constraint of chord.

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

2xA+xB/A=constant (1)

Here,xA is the length of cord connecting block A and xB/A is the length of cord connecting block B relative to block A.

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

2vA+vB/A=constant (2)

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

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

2aA+aB/A=0aB/A=2aA (3)

Here, aA is the acceleration of block A and aB/A acceleration of block B relative to block A.

Find the equation of acceleration of block B in terms of acceleration of block A.

aB=aA+aB/A

Here, aB is the acceleration of block B.

Substitute 2aA for aB/A.

aB=aA+(2aA)=aA (4)

First of all determine whether the blocks will move for the given value of P(40N).

Sketch the free body diagram of block B as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 12.1, Problem 12.19P , additional homework tip  1

Refer Figure (1),

Consider equilibrium along y-axis .

ΣFy=0NABWBcos25°=0NAB=WBcos25°

Here, NAB is the normal force on block B from block A.

Substitute mBg for WB.

NAB=mBgcos25°

Write the equation of frictional force (FAB) on the block B from A.

FAB=μsNAB

Substitute 0.20 for μs and mBgcos25° for NAB.

FAB=0.20mBgcos25°

Consider equilibrium along x-axis.

ΣFx=0T+FAB+WBsin25°=0T=FAB+WBsin25°

Substitute 0.20mBgcos25° for FAB and mBg for WB.

T=0.20mBgcos25°+mBgsin25°=mBg(0.20cos25°+sin25°)

Substitute 8 kg for mB and 9.81m/s2 for g.

T=(8)(9.81)(0.20cos25°+sin25°)=47.39249N (5)

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

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 12.1, Problem 12.19P , additional homework tip  2

Refer Figure (2).

Consider equilibrium along y-axis.

ΣFy=0NANABWAcos25°+Psin25°=0NA=NAB+WAcos25°Psin25°

Here, NA is the normal force on block A.

Substitute mBgcos25° for NAB and mAg for WA.

NA=mBgcos25°+mAgcos25°Psin25°=gcos25°(mA+mB)Psin25°

Write the equation of frictional force FA on the block A.

FA=μsNA

Substitute 0.20 for μs and gcosθ(mA+mB) for NA.

FA=0.20[gcosθ(mA+mB)Psin25°]

Consider equilibrium along x-axis.

ΣFx=0TFAFAB+WAsinθ+Pcos25°=0T=WAsinθFAFAB+Pcos25°

Substitute 0.20[gcosθ(mA+mB)Psin25°] for FA, 0.20mBgcos25° for FAB, and mAg for WA.

T=mAgsin25°0.20[gcos25°(mA+mB)Psin25°]0.20mBgcos25°+Pcos25°=mAgsin25°0.20gcos25°(mA+2mB)+P(0.20sin25°+cos25°)

Substitute 40 kg for mA, 9.81m/s2 for g, and 8 kg for mB.

T=(40)(9.81)sin25°0.20(9.81)cos25°(40+2×8)+P(0.20sin25°+cos25°)=165.8354199.57785+0.99083P=66.25756+0.99083P (6)

Find the force P for impending motion.

Equate Equation (5) and (6).

47.39249=66.25756+0.99083PP=19.04N

The force P of impending motion is less than the 40N, the blocks will move. Therefore apply Newton’s law of motion.

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

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 12.1, Problem 12.19P , additional homework tip  3

Refer Figure (3),

Consider equilibrium along y-axis .

ΣFy=0NABWBcos25°=0NAB=WBcos25°

Substitute mBg for WB.

NAB=mBgcos25°

Write the equation of frictional force (FAB) on the block B from A.

FAB=μkNAB

Substitute 0.15 for μk and mBgcos25° for NAB.

FAB=0.15mBgcos25°

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

ΣFx=mBaBT+FAB+WBsin25°=mBaBT=FAB+WBsin25°mBaB

Substitute 0.15mBgcos25° for FAB and mBg for WB.

T=0.15mBgcos25°+mBgsin25°mBaB=mB(0.15gcos25°+gsin25°aB)

Substitute 8 kg for mB and 9.81m/s2 for g.

T=mB(0.15gcos25°+gsin25°aB)=(8)(0.15×9.81cos25°+9.81sin25°aB)=8(5.47952aB) (7)

Sketch the free body diagram and kinetic diagram of block A as shown in Figure 4.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 12.1, Problem 12.19P , additional homework tip  4

Refer Figure (4).

Consider equilibrium along y-axis.

ΣFy=0NANABWAcos25°+Psin25°=0NA=NAB+WAcos25°Psin25°

Substitute mBgcos25° for NAB and mAg for WA.

NA=mBgcos25°+mAgcosθPsin25°=gcos25°(mA+mB)Psin25°

Write the equation of frictional force FA on the block A.

FA=μsNA

Substitute 0.15 for μk and gcos25°(mA+mB)Psin25° for NA.

FA=0.15[gcos25°(mA+mB)Psin25°]

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

ΣFx=mAaATFAFAB+WAsin25°+Pcos25°=0T=WAsin25°FAFAB+Pcos25°mAaA

Substitute 0.15[gcos25°(mA+mB)Psin25°] for FA, 0.15mBgcos25° for FAB, mAg for WA and aB for aA.

T=mAgsin25°0.15[gcos25°(mA+mB)Psin25°]0.15mAgcos25°+Pcos25°mA(aB)=mAgsin25°0.15gcos25°(mA+2mB)+0.15Psin25°+Pcos25°+mAaB

Substitute 40 kg for mA, 8 kg for mB, 40 N for P, and 9.81m/s2 for g.

T=(40)(9.81)sin25°0.15(9.81)cos25°(40+2×8)+0.15(40)sin25°+(40)cos25°+(40)aB=165.8354174.68339+2.53571+36.25231+40aB=130+40aB (8)

Find the acceleration of block B (aB).

Equate equation (7) and (8).

8(5.47952aB)=130+40aB43.836168aB=130+40aB48aB=86.16384aB=1.795m/s2

Negative sign indicates the motion of block B in opposite to x-axis.

Thus, the acceleration of block B is 1.795m/s2_.

(b)

Expert Solution
Check Mark
To determine

Find the tension in the cord.

Answer to Problem 12.19P

The tension in the cord is 58.2N_.

Explanation of Solution

Calculation:

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

T=130+40aB

Substitute 1.795m/s2 for aB.

T=130+40(1.795)=58.2N

Thus, the tension in the cord is 58.2N_.

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Chapter 12 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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