VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 12.1, Problem 12.22P

To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are μs = 0.40 and μk = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s.

Fig. P12.22

Chapter 12.1, Problem 12.22P, To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and

(a)

Expert Solution
Check Mark
To determine

Find the smallest acceleration of the truck which will cause the stack of plywood to slide.

Answer to Problem 12.22P

The smallest acceleration of the truck which will cause the stack of plywood to slide is 0.309m/s2_.

Explanation of Solution

Given information:

The coefficients of static friction (μs) between the bottom sheet of plywood and the bed is 0.40.

The coefficients of static friction (μk) between the bottom sheet of plywood and the bed is 0.30.

The tilting angle (θ) of bed of truck is 20°.

The relative distance of plywood with respect to truck (xP/T) is 2 m.

Calculation:

Write the equation of Weight of plywood (WP).

WP=mPg

Here, mP is the mass of plywood and g is the acceleration of gravity.

Sketch the free body diagram and kinetic diagram of plywood as shown in Figure (1).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 12.1, Problem 12.22P , additional homework tip  1

Refer Figure (1).

Apply the Newton’s law of equation along y-axis.

ΣFy=mPaTsin20°N1WPcos20°=mPaTsin20°

Substitute mPg for WP.

N1mPgcos20°=mPaTsin20°N1=mP(gcos20°aTsin20°)

Write the equation of friction force (F1).

F1=μsN1

Here, N1 is the normal force on the plywood.

Substitute 0.40 for μs.

F1=0.40N1

Substitute mP(gcos20°aTsin20°) for N1.

F1=0.40mP(gcos20°aTsin20°) (1)

Apply the Newton’s law of equation along x-axis.

ΣFx=maxF1WPsin20°=mPaTcos20°

Substitute mPg for WP.

F1(mPg)sin20°=mPaTcos20°F1=mP(gsin20°+aTcos20°) (2)

Find the smallest acceleration of the truck which will cause the stack of plywood to slide.

Equate Equation (1) and (2).

0.40mP(gcos20°aTsin20°)=mP(gsin20°+aTcos20°)0.40gcos20°0.40aTsin20°=gsin20°+aTcos20°0.40gcos20°gsin20°=aTcos20°+0.40aTsin20°0.0339g=1.0765aT

Substitute 9.81m/s2 for g.

0.0339(9.81)=1.0765aTaT=0.309m/s2

Thus, the smallest acceleration of the truck which will cause the stack of plywood to slide is 0.309m/s2_.

(b)

Expert Solution
Check Mark
To determine

Find the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s.

Answer to Problem 12.22P

The acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s is 4.17m/s2_.

Explanation of Solution

Calculation:

The velocity of plywood relative to truck is zero.

Find the acceleration of the plywood relative to the truck (aP/T) using equation of kinematics.

xP/T=(vP/T)t+12aP/Tt2

Here, vP/T is the velocity of plywood relative to truck.

Substitute 0 for vP/T 2 m for xP/T, and 0.9 s for t.

2=0+12aP/T(0.9)2aP/T=4.94m/s2

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

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 12.1, Problem 12.22P , additional homework tip  2

Refer Figure (2).

Apply the Newton’s law of equation along y-axis.

ΣFy=mPayN2WPcos20°=mPaTsin20°

Substitute mPg for WP.

N2(mPg)cos20°=mPaTsin20°N2=mP(gcos20°aTsin20°) (3)

Here, N2 is the normal force on truck.

Apply the Newton’s law of equation along x-axis.

ΣFx=ΣmaxF2WPsin20°=mPaTcos20°mPaP/T

Substitute mPg for WP.

F2(mPg)sin20°=mPaTcos20°mPaP/TF2=mP(gsin20°+aTcos20°aP/T) (4)

Here, F2 is the frictional force on truck.

Write the equation of frictional force on truck:

F2=μkN2

Substitute 0.30 for μk and mP(gcos20°aTsin20°) for N2.

F2=0.30mP(gcos20°aTsin20°) (5)

Find the acceleration of the truck (aT) which causes corner A of the stack to reach the end of the bed in 0.9 s

Equate Equation (4) and (5),

mP(gsin20°+aTcos20°aP/T)=0.30mP(gcos20°aTsin20°)aTcos20°+0.30aTsin20°=0.30gcos20°gsin20°+aP/TaT(cos20°+0.30sin20°)=0.30gcos20°gsin20°+aP/TaT=g(0.30cos20°sin20°)+aP/Tcos20°+0.30sin20°

Substitute 9.81m/s2 for g, 4.94m/s2 for aP/T

aT=9.81(0.30cos20°sin20°)+4.94cos20°+0.30sin20°=4.351.042=4.17m/s2

Thus, the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s is 4.17m/s2_.

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