All 20 Chapters Covered
SOLUTIONMANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5 Impending motion to left
E
1 1
f f
A B
G
Fcr
F
D C cr
Facc
Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric-
tion DE and BE to find the point of concurrency at E for impending motion to the left. The
critical angle is θcr. Resolve force F into components Facc and Fcr. Facc is related to mass
and acceleration. Pin accelerates to left for any angle 0 < θ < θcr. When θ > θcr, no
magnitude of F will move the pin.
Impending motion to right
E E
1 1
f f
A B
G
d
F F
cr
D C cr
acc
F
Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric-
tion AE ′ and CE ′ to find the point of concurrency at E ′ for impending motion to the left.
The critical angle is θc′r . Resolve force F ′ into components Fa′ cc and Fc′r . Fa′ cc is related
to mass and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θc′r . When θ ′ > θc′r ,
no mag- nitude of F ′ will move the pin.
The intent of the question is to get the student to draw and understand the free body in
order to recognize what it teaches. The graphic approach accomplishes this quickly. It is im-
portant to point out that this understanding enables a mathematical model to be constructed,
and that there are two of them.
This is the simplest problem in mechanical engineering. Using it is a good way to begin a
course.
What is the role of pin diameter d?
Yes, changing the sense of F changes the response.
,2 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering
Design
1-6
(a) y Fy = −F − f N cos θ + N sin θ = 0 (1)
T
F
Fx = f N sin θ + N cos θ =0
T − r
r x F = N (sin θ − f cos θ) Ans.
N
T = Nr ( f sin θ + cos θ)
fN
Combining
1 + f tan θ
T = Fr
tan θ = KFr Ans. (2)
− f
(b) If T → ∞ detent self-locking tan θ − f = 0 ∴ θcr = tan−1 f Ans.
(Friction is fully developed.)
Check: If F = 10 lbf, f = 0.20, θ = 45◦, r = 2 in
10
N= = 17.68 lbf
−0.20 cos 45◦ + sin
45◦
T
= 17.28(0.20 sin 45◦ + cos 45◦) = 15 lbf
r
f N = 0.20(17.28) = 3.54 lbf
θcr = tan−1 f = tan−1(0.20) = 11.31◦
11.31° < θ < 90°
1-7
(a) F = F0 + k(0) =
F0 T1 = F0r
Ans.
(b) When teeth are about to clear F = F0 + kx 2
From Prob. 1-6 f tan θ + 1
T2 = Fr
tan θ − f
( F0 + kx 2 )( f tan θ + 1)
T2 = r Ans.
tan θ f
−
1-8
Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦, f = 0.25, xi = 0, x f = 0.2
Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.
, Chapter 1 3
From Eq. (1) of Prob. 1-6
N= F
Ni = — f cos θ + sin θ = 13.49 lbf Ans.
10
−0.25 cos 60◦ + sin
60◦
10.5
Nf 13.49 14.17 lbf Ans.
10 =
=
From Eq. (2) of Prob. 1-6
1 + f tan θ 1 + 0.25 tan 60◦
K
= tan = 0.967 Ans.
= −0.25
tan θ − f 60◦
Ti = 0.967(10)(2) = 19.33 lbf · in
Tf = 0.967(10.5)(2) = 20.31 lbf · in
1-9
(a) Point vehicles
v
x
cars v 42.1v − v2
Q= =
hour x
= 0.324
Seek stationary point maximum
dQ 42.1 − 2v
∴ v* = 21.05 mph
=0=
dv 0.324
42.1(21.05) − 21.052
Q*
= = 1367.6 cars/h Ans.
0.324
(b) v
l x l
2 2
Q= v 0.324 −1
x l
+l +
= v(42.1) − v
v2
Maximize Q with l = 10/5280 mi
SOLUTIONMANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5 Impending motion to left
E
1 1
f f
A B
G
Fcr
F
D C cr
Facc
Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric-
tion DE and BE to find the point of concurrency at E for impending motion to the left. The
critical angle is θcr. Resolve force F into components Facc and Fcr. Facc is related to mass
and acceleration. Pin accelerates to left for any angle 0 < θ < θcr. When θ > θcr, no
magnitude of F will move the pin.
Impending motion to right
E E
1 1
f f
A B
G
d
F F
cr
D C cr
acc
F
Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric-
tion AE ′ and CE ′ to find the point of concurrency at E ′ for impending motion to the left.
The critical angle is θc′r . Resolve force F ′ into components Fa′ cc and Fc′r . Fa′ cc is related
to mass and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θc′r . When θ ′ > θc′r ,
no mag- nitude of F ′ will move the pin.
The intent of the question is to get the student to draw and understand the free body in
order to recognize what it teaches. The graphic approach accomplishes this quickly. It is im-
portant to point out that this understanding enables a mathematical model to be constructed,
and that there are two of them.
This is the simplest problem in mechanical engineering. Using it is a good way to begin a
course.
What is the role of pin diameter d?
Yes, changing the sense of F changes the response.
,2 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering
Design
1-6
(a) y Fy = −F − f N cos θ + N sin θ = 0 (1)
T
F
Fx = f N sin θ + N cos θ =0
T − r
r x F = N (sin θ − f cos θ) Ans.
N
T = Nr ( f sin θ + cos θ)
fN
Combining
1 + f tan θ
T = Fr
tan θ = KFr Ans. (2)
− f
(b) If T → ∞ detent self-locking tan θ − f = 0 ∴ θcr = tan−1 f Ans.
(Friction is fully developed.)
Check: If F = 10 lbf, f = 0.20, θ = 45◦, r = 2 in
10
N= = 17.68 lbf
−0.20 cos 45◦ + sin
45◦
T
= 17.28(0.20 sin 45◦ + cos 45◦) = 15 lbf
r
f N = 0.20(17.28) = 3.54 lbf
θcr = tan−1 f = tan−1(0.20) = 11.31◦
11.31° < θ < 90°
1-7
(a) F = F0 + k(0) =
F0 T1 = F0r
Ans.
(b) When teeth are about to clear F = F0 + kx 2
From Prob. 1-6 f tan θ + 1
T2 = Fr
tan θ − f
( F0 + kx 2 )( f tan θ + 1)
T2 = r Ans.
tan θ f
−
1-8
Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦, f = 0.25, xi = 0, x f = 0.2
Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.
, Chapter 1 3
From Eq. (1) of Prob. 1-6
N= F
Ni = — f cos θ + sin θ = 13.49 lbf Ans.
10
−0.25 cos 60◦ + sin
60◦
10.5
Nf 13.49 14.17 lbf Ans.
10 =
=
From Eq. (2) of Prob. 1-6
1 + f tan θ 1 + 0.25 tan 60◦
K
= tan = 0.967 Ans.
= −0.25
tan θ − f 60◦
Ti = 0.967(10)(2) = 19.33 lbf · in
Tf = 0.967(10.5)(2) = 20.31 lbf · in
1-9
(a) Point vehicles
v
x
cars v 42.1v − v2
Q= =
hour x
= 0.324
Seek stationary point maximum
dQ 42.1 − 2v
∴ v* = 21.05 mph
=0=
dv 0.324
42.1(21.05) − 21.052
Q*
= = 1367.6 cars/h Ans.
0.324
(b) v
l x l
2 2
Q= v 0.324 −1
x l
+l +
= v(42.1) − v
v2
Maximize Q with l = 10/5280 mi
