Sample Problem 17.3The mass center of the 10-kg sliding door in Fig. (a) is located at G. The door

y

x

1 m

1.5 m

1 m

h

P

G

A B

(a)

is supported on the horizontal rail by sliders at A and B. The coefficient of staticas well as kinetic friction is 0.4 at A and 0.3 at B. The door was at rest beforethe horizontal force P = 60 N was applied. (1) Find the maximum value of hfor which the door will slide to the right without tipping and the correspondingacceleration of the door. (2) If h = 1.5 m, find all forces acting on the door, andcalculate its acceleration.

SolutionPart 1

The free-body and mass-acceleration diagrams for the door, shown in Fig. (b), aredescribed in detail as follows.

Free-body diagram The FBD contains the following forces: the 10(9.8) = 98 Nweight acting at G, the applied force P, the normal force NA, and the frictionforce FA =μA NA = 0.4NA. Because the door starts from rest, its velocity will bedirected to the right (i.e., in the same direction as the acceleration), which meansthat FA is directed to the left. There is no normal force and, therefore, no frictionforce at B, because the problem statement implies that the door is sliding to theright in a state of impending tipping about A.

Mass-acceleration diagram The MAD contains only the inertia vector of magni-tude ma acting at G. There is no inertia couple because the door is translating(α = 0).

1 m

h

P = 60 N

G

A B

(b)

1.5 m

G

A B

MAD

FA = 0.4NA

NA

98 N

FBD

10a N

=

Inspection of Fig. (b) reveals that the unknowns are NA and h on the FBD,and a on the MAD. These three unknowns can be computed by deriving andsolving any three independent equations of motion. When employing the FBD-MAD technique, remember that (1) the resultant force on the FBD can be equatedto the inertia vector ma on the MAD, and (2) the resultant moment about anypoint on the FBD can be equated to the resultant moment about the same point onthe MAD.

377

Equating the x- and y-components of the forces on the FBD to the correspond-ing components of the inertia vector, we obtain

�Fy = may +�⏐ NA − 98 = 0

NA = 98 N

and

�Fx = max −→+ 60 − 0.4NA = 10a

60 − 0.4(98) = 10a

which givesa = 2.08 m/s2 Answer

The third independent equation is a moment equation about any point. If wechoose point A as the moment center, the resultant moment on the FBD is equatedto the resultant moment on the MAD:

(�MA)FBD = (�MA)MAD + 60h − 10(9.8)(1.5) = 10a(1.5)

With a = 2.08 m/s2, we geth = 2.97 m Answer

Part 2

The FBD and MAD for h = 1.5 m are shown in Fig. (c). The details of thesediagrams are as follows.

Free-body diagram The FBD contains the 98-N weight, the applied force P =60 N passing through G, the normal forces NA and NB , and the friction forces FA

and FB . From the solution of Part 1 we know that the door will be sliding to theright while maintaining contact at both A and B. The friction forces, determinedby the kinetic coefficients of friction, are directed to the left—that is, opposite tothe motion.

Mass-acceleration diagram Because the door is sliding to the right without rotat-ing, the MAD contains only the inertia vector acting through G.

1 m

h = 1.5 m

P = 60 NG

A B

(c)

1.5 m

G

A B

MAD

FA = 0.4NA

NA

98 N

FBD

10 a N=

NB

1 m

FB = 0.3NB

378

The number of unknowns in the FBD and MAD is three: NA, NB , and a,which can be found from any three independent equations of motion. One suchset of equations is given below; their validity can be determined by referring tothe FBD and MAD in Fig. (c).

�Fy = may +�⏐ NA + NB − 98 = 0 (a)

�MG = 0 + NB(1)+ 0.3NB(1.5)− NA(1)+ 0.4NA(1.5) = 0 (b)

�Fx = max −→+ 60 − 0.4NA − 0.3NB = 10 a (c)

Solving Eqs. (a), (b), and (c) gives

NA = 76.85 N NB = 21.2 N a = 2.29 m/s2 Answer

The positive values of NA and NB confirm that the door will not tip.

Sample Problem 17.4The homogeneous bar in Fig. (a) has mass m and length L. The bar, which is free

Ox

y

L

(a)

θ

to rotate in the vertical plane about a pin at O, is released from rest in the positionθ = 0. Find the angular acceleration α when θ = 60◦.

SolutionFigure (b) shows the FBD and the MAD of the bar when θ = 60◦. The FBD con-tains the weight W of the bar, acting at its mass center G (located at the midpointof the bar) and the components of the pin reaction at O. In the MAD, the inertiacouple Iα was drawn assuming that α is clockwise, and using I = mL2/12 fromTable 17.1. The components of the inertia vector ma were found by noting thatthe path of G is a circle centered at O. Therefore, the normal and tangential com-ponents of a are an = (L/2)ω2 and at = (L/2)α. The direction of an is toward O,regardless of the direction of ω. The direction of at is consistent with the assumeddirection of α.

O

G

W = mg

Ox

Oy

60°

FBD

O

L2 G

60°

MAD

mat = m αL2

man = m L ω2L2

Iα = mL2

12=

(b)

L2

α

αα

ω

379

We note that there are a total of four unknowns in Fig. (b): Ox , Oy , α, and ω.Because there are only three independent equations of motion, we will not beable to determine all unknowns using only the FBD and the MAD. The reason forthis is that ω depends on the history of motion: ω = ∫

α dt + C . Therefore, theequations of motion at a specific position of the bar will not determine the angu-lar velocity in that position. However, inspection of the FBD and MAD revealsthat it is possible to determine the angular acceleration α, because it is the onlyunknown that appears in the moment equation when O is used as the momentcenter. Referring to the diagrams in Fig. (b), this moment equation is

(�MO)FBD = (�MO)MAD

+ mgL

2cos 60◦ = mL2

12α +

(m

L

2α

)L

2= mL2

3α (a)

from which we find that

α = 3g

2Lcos 60◦ = 0.750

g

LAnswer (b)

Because the acceleration of O is zero, the above moment equation could havealso been obtained from Eq. (17.14): �MO = IOα, where

IO = I + md2 = mL2

12+ m

(L

2

)2

= mL2

3(c)

from the parallel-axis theorem. We now see that �MO = IOα will yield anequation that is identical to Eq. (a).

Sample Problem 17.5The body shown in Fig. (a) consists of the homogeneous slender bar 1 that is

= 30°

L = 800 mm

m1 = 30 kg

= 1.2 rad/s

m2 = 80 kg

(a)

1

2

R = 200 mm

Oθ

ω

rigidly connected to the homogeneous sphere 2. The body is rotating in the verti-cal plane about the pin at O. When the body is in the position where θ = 30◦, itsangular velocity is ω = 1.2 rad/s clockwise. At this instant, determine the angularacceleration α and the magnitude of the pin reaction at O.

SolutionThe FBD and MAD of the body in the position θ = 30◦ are shown in Fig. (b). Inthese diagrams, the bar and the sphere are treated as separate entities, each with itsown inertia couple and inertia vector. (An equivalent form of the MAD would beobtained by showing the inertia couple and inertia vector for the assembly; referto Fig. 17.13.) Details of the diagrams are described in the following.

Free-body diagram The forces On and Ot are the components of the pin reactionrelative to the n- and t-axes shown in the figure. The weights W1 and W2 of thebar and sphere, respectively, act at their mass centers G1 and G2. The distances

380

= 30°

r2 = 1.0 m

(b)

30°

r1 = 0.4 m

30°

W1 = 30(9.81) N W

2 = 80(9.81) N

G1

OOn

Ot

n

t G2

FBD

r2 = 1.0 m

r1 = 0.4 m

G1

O

G2

MAD

m1 r1 2 = 17.28 N

I1 = 1.600 N·m

m1 r1 = 12.00 N

m2 r2 2 = 115.2 N m

2 r2 = 80.00 N

I2 = 1.280 N·m

=

θ α

α

α

α

α

α

α

α

ω

ω

r1 = 0.4 m and r2 = 1.0 m, measured from O to the mass centers, are deducedfrom the dimensions in Fig. (a).

Mass-acceleration diagram In the MAD, we assume that the angular accelera-tion α, measured in rad/s2, is clockwise. Using the fact that the body rotates aboutthe fixed point O, kinematic analysis enables us to express the accelerations of G1

and G2 in terms of α and ω of the body. The inertia terms that appear in the MADhave been computed in the following manner.For the slender bar:

I1α = m1L2

12α = 30(0.8)2

12α = 1.600α N · m

m1r1ω2 = 30(0.4)(1.2)2 = 17.28 N

m1r1α = 30(0.4)α = 12.00α N

For the sphere:

I2α = 2

5m2 R2α = 2

5(80)(0.2)2α = 1.280α N · m

m2r2ω2 = 80(1.0)(1.2)2 = 115.2 N

m2r2α = 80(1.0)α = 80.00α N

In the MAD, the directions of the tangential components of the inertia vectors(those containing α) are consistent with the assumed clockwise direction of α.The normal components of the inertia vectors (those containing ω2) are directedtoward the center of rotation O, regardless of the direction of ω.

From Fig. (b) we see that there are two unknowns in the FBD (On and Ot ) andone unknown (α) in the MAD. Therefore, all that remains is to derive and solvethe three independent equations of motion for the unknowns.

381

Equating moments about O on the FBD and the MAD in Fig. (b), we obtain

(�MO)FBD = (�MO)MAD

+ 30(9.8)(0.4) cos 30◦ + 80(9.8)(1.0) cos 30◦

= 1.600α + (12.00α)(0.4)+ 1.280α + (80.00α)(1.0)

= 87.68α

from which we find thatα = 8.905 rad/s2 Answer

Because the acceleration of point O is zero, this result could also have beenderived using the special case �MO = IOα.

Using α = 8.914 rad/s2 and referring to Fig. (b), the force equations in thet and n directions give

�Ft = mat

+↙ Ot + 30(9.8) cos 30◦ + 80(9.8) cos 30◦ = 12.00(8.905)+ 80.00(8.905)

Ot = −114.3 N

and

�Fn = man

↖+ On − 30(9.8) sin 30◦ − 80(9.8) sin 30◦ = 17.28 + 115.2

On = 671.5 N

Therefore, the magnitude of the pin reaction at O is

O =√

O2t + O2

n =√(−114.3)2 + (671.5)2 = 681.2 N Answer

Sample Problem 17.6The cable connected to block B in Fig. (a) is wound tightly around disk A, which is

B mB = 20 kg

y

x

R = 500 mm

mA = 60 kg

k = 400 mm

A

G

(a)

free to rotate about the axle at its mass center G. The masses of A and B are 60 kgand 20 kg, respectively, and k = 400 mm for the disk. Determine the angularacceleration of A and the tension in the cable.

SolutionThe free-body and mass-acceleration diagrams of the system are shown inFig. (b). The FBD contains the weights WA = 60(9.8) = 588 N and WB =20(9.8) = 196 N together with the unknown pin reactions at G. The tension inthe cable, being an internal force, does appear on this FBD.

The MAD displays the inertia couple of the disk and the inertia vector of theblock. There is no inertia vector of the disk because its mass center G is stationary.The angular acceleration α of the disk is assumed to be directed clockwise. Thecorresponding inertia couple of the disk is

Iα = mk2α = 60(0.4)2α = 9.600α N · m

382

G

WA = 588 N

Gx

0.5 m

Gy

FBD

WB = 196 N

G0.5

m

I = 9.600 N·m

MAD

mBaB = 10 N

=

(b)

α

α α

which also is directed clockwise. Because the cable does not slip on the disk, theacceleration of the block is aB = Rα, which results in the inertia vector

m BaB = m B(Rα) = 20(0.5α) = 10α N⏐�

There are three unknowns on the FBD and the MAD: two components of thepin reaction at G and the angular acceleration α of the disk. Since the numberof independent equations available from the FBD and MAD is also three, all theunknowns can be computed.

The angular acceleration α can be found by equating the resultant momentsabout G in the FBD and the MAD.

(�MG)FBD = (�MG)MAD + 196(0.5) = 9.600α + 10α(0.5)

α = 6.712 rad/s2 Answer

To find the tension in the cable, we analyze the block separately (the disk couldalso be used). The FBD and MAD for the block are shown in Fig. (c), where T isthe cable tension. Summing forces in the y-direction yields

�Fy = may +⏐� 196 − T = 10α = 10(6.712)

T = 129.0 N Answer

WB = 196 N

FBD

T

= MAD

mBaB = 10 N

(c)

α

383

Sample Problem 17.7The 40-kg unbalanced wheel in Fig. (a) is rolling without slipping under the actionof a counterclockwise couple C0 = 20 N · m. When the wheel is in the positionshown, its angular velocity is ω = 2 rad/s, clockwise. For this position, calculatethe angular acceleration α and the forces exerted on the wheel at C by the roughhorizontal plane. The radius of gyration of the wheel about its mass center G isk = 200 mm.

R = 250 mm

O

y

x

120 mm

GC0 = 20 N·m

ω = 2 rad/s

(a)

C C C

ω

SolutionThe free-body and mass-acceleration diagrams for the wheel, shown in Fig. (b),were constructed as follows.

Free-body diagram The FBD consists of the applied couple C0, the weightW = 40(9.8)= 392 N, and the normal and friction forces that act at the contactpoint C, denoted by NC and FC , respectively. Observe that FC has been assumedto be directed to the right.

Mass-acceleration diagram In the MAD of Fig. (b) the angular acceleration α,measured in rad/s2, has been assumed to be clockwise. The corresponding inertiacouple shown on this diagram is

Iα = mk2α = 40(0.200)2α = 1.600α N · m

O

0.12 m

G

FBD

W = 392 N

C FC

NC

G

MAD

C

0.12 m

R = 0.25 m

C0 = 20 N.m

(b)

ma–y = 4.80 N

ma–x = (10.00 – 19.20) N

I = 1.600 N·m

O=

α α

α

α

384

Because the wheel does not slip, the acceleration of its center O is aO=Rα= 0.250α m/s2, directed to the right. Applying the relative accelerationequation between G and O, we obtain (the units of each term are m/s2)

a = aG = aG/O

ax

ay

G

0.120

O0.120 m0.250 0.120 2 = 0.120(2)2

= 0.480

aO +

α

α

α

ω

from which we find ax = 0.250α − 0.480 m/s2 and ay = 0.120α m/s2. Multi-plying these results by m = 40 kg, the components of the inertia vector becomemax = (10.00α − 19.20) N, directed to the right, and may = 4.80α N, directeddownward.

The FBD and MAD in Fig. (b) now contain only three unknowns: NC , FC ,and α, which can be found using any three independent equations of motion.

Because NC and FC act at C, it is convenient to use that point as a momentcenter, the corresponding moment equation being

(�MC)FBD = (�MC)MAD

+ − 20 + 392(0.120) = 1.600α + 0.250(10.00α − 19.20)+ 0.120(4.80α)

The solution to this equation is

α = 6.820 rad/s2 Answer

Because α is positive, its direction is clockwise, as assumed.The forces at C can now be found from force equations of motion:

�Fx = max −→+ FC = 10.00α − 19.20 = 10.00(6.820)− 19.20

and

�Fy = may +⏐� 392 − NC = 4.80α = 4.80(6.820)

which yield

FC = 49.0 N and NC = 359.3 N Answer

Because each force is positive, it is directed as shown in the FBD.

385

Sample Problem 17.8Figure (a) shows a 10-kg homogeneous disk of radius 0.2 m. The disk is at restbefore the horizontal force P = 60 N is applied to its mass center G. The coeffi-cients of static and kinetic friction for the surfaces in contact are 0.20 and 0.15,respectively. Determine the angular acceleration of the disk and the accelerationof G after the force is applied.

SolutionTwo motions of the disk are possible: rolling without slipping, and rolling withslipping. We will solve the problem by assuming that the disk rolls without slip-

R = 0.2 m

P = 60 NG

s = 0.20

k = 0.15

y

x

(a)

μμ

ping. This assumption will then be checked by comparing the required frictionforce with its maximum static value.

The free-body diagram (FBD) and the mass-acceleration diagram (MAD)based on the no-slip assumption are shown in Fig. (b). The FBD contains the10(9.8)= 98 N weight, the 60 N applied force, the normal force NA, and the fric-tion force F, assumed acting to the left. The MAD contains the inertia coupleand inertia vector, where the angular acceleration α, measured in rad/s2, has beenassumed to be clockwise. The values of Iα and ma were computed as follows.

Iα = m R2

2α = 10(0.2)2

2α = 0.2α N · m

ma = m Rα = 10(0.2)α = 2α N

P = 60 N

G

FBD

98 N

0.2 m

C

NA

F

G

MAD

C

Iα = 0.2α

ma = 2 N

0.2 m=

(b)

αα

α

Note that a = Rα is a valid kinematic equation because the disk is assumed tobe rolling without slipping. There are a total of three unknowns in the FBD andMAD: F, N, and α, which can be computed using any three independent equationsof motion.

A convenient solution is to first equate the resultant moment about C on theFBD to the resultant moment about C on the MAD and then utilize the forceequations of motion.

386

(�MC)FBD = (�MC)MAD+ 60(0.2) = 0.2α + 2α(0.2)

α = 20 rad/s2

�Fx = max −→+ 60 − F = 2α = 2(20)

F = 20 N

�Fy = may +�⏐ NA − 98 = 0

NA = 98 N

Because α, F, and NA are all positive, their directions are as shown in Fig. (b).Next we note that the maximum possible static friction force is Fmax =

μs NA = 0.20(98) = 19.6. The friction force required for rolling without slip-ping is, according to our solution, F = 20 N. Because F > Fmax, we concludethat the disk does slip, and we must reformulate the problem.

The FBD and MAD for the case where the disk rolls and slips simultaneouslyare shown in Fig. (c). The friction force F in the FBD has been set equal to itskinetic value, μk N . This force must be shown acting to the left in order to opposeslipping. The inertia couple Iα in the MAD is identical to that used in Fig. (b).However, the important difference here is that the magnitude of the inertia vectoris now ma = 10a N, where a is measured in m/s2. Because the disk is slipping,the kinematic constraint a = Rα does not apply. Once again we see that there arethree unknowns on the FBD and the MAD, except that now the unknowns are N,α, and a.

P = 60 N

G

FBD

98 N

0.2 m

C

NA

F = 0.15 NA

G

MAD

C

=

(c)

0.5 ft

Iα = 0.2 α αα

ma = 10a N

The three unknowns can be calculated as follows (of course, any other threeindependent equations could also be used):

�Fy = may +�⏐ NA − 98 = 0 N = 98 N (a)

�Fx = max −→+ 60 − 0.15NA = 10a (b)

�MG = Iα + 0.2(0.15NA) = 0.2α (c)

Substituting N = 50 lb from Eq. (a) into Eqs. (b) and (c) yields

a = 4.53 m/s2 and α = 14.7 rad/s2 Answer

387

Sample Problem 17.9A homogeneous slender bar AB of mass m and length L is released from rest in theposition shown in Fig. (a). Determine the acceleration of end A, the reaction at A,and the angular acceleration of the bar immediately after the release. Assume thatthe horizontal plane is frictionless.

B

m

60°

(a)

A A

A

L

Solution

L

B

60°A

N

G

mg2

B

60°A

G

60°

maA I = mL212

Lm

y

x

MADFBD

(b)

L 2

2=

α

α

α

The free-body diagram (FBD) and mass-acceleration diagram (MAD) of the barat the instant after the release are shown in Fig. (b). The FBD contains theweight of the bar, mg, and the vertical reaction N. The MAD contains the inertiacouple, Iα, and the inertia vector, ma. The latter consists of components maA andm(L/2)α, which were obtained from kinematics as follows:

a = aG = aA

A

G

60°L2

L2

L2

2 = 0+ aG/A

aA

α

α

ω(a)

In Eq. (a), the senses of aA and α were assumed to be to the left and clockwise,respectively. The angular velocity ω is zero because the bar has just been releasedfrom rest in the position being considered. Multiplying the right-hand side ofEq. (a) by the mass m and placing the results at G gives the components of theinertia vector shown in the MAD of Fig. (b).

388

Inspection of Fig. (b) reveals that there are a total of three unknowns: N, aA,and α. Therefore, the solution can be obtained by deriving and solving any threeindependent equations of motion.

Equating moments about A on the FBD and MAD in Fig. (b) yields

(�MA)FBD = (�MA)MAD

+ mg

(L

2cos 60◦

)= mL2

12α + m

L

2α

(L

2

)− maA

(L

2sin 60◦

)

which, on simplification, becomes

aA = 0.7698Lα − 0.5774g (b)

Referring again to Fig. (b), the force equation for the horizontal direction becomes

�Fx = max −→+ 0 = −maA + mL

2α sin 60◦

which reduces to

aA = 0.4330Lα (c)

Solving Eqs. (b) and (c) simultaneously yields

aA = 0.742g and α = 1.714g

LAnswer

Using the diagrams in Fig. (b), the force equation for the vertical direction is

�Fy = may +�⏐ − mg + N = −m

L

2α cos 60◦ (d)

Substituting the expression for α found above, and solving for N, gives

N = 0.572 mg Answer

It must be emphasized that the values obtained for N, α, and aA are valid onlyat the instant after the release. Each of these variables will vary throughout thesubsequent motion of the bar. However, it is interesting to note that there is nevera horizontal force acting on the bar because the plane is frictionless. Therefore,the path followed by the mass center G will be a vertical straight line.

389