All 20 Chapters Covered
SOLUTION MANUAL
, 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
cr F
D C cr
Facc
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)
F T
Fx = f N sin θ + N cos θ − = 0
r
T
r x F = N (sin θ − f cos θ ) Ans.
T = Nr ( f sin θ + cos θ )
N
fN
Combining
1 + f tan θ
T = Fr = KFr Ans. (2)
tan θ − 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 + kx2
From Prob. 1-6
f tan θ + 1
T2 = Fr
tan θ − f
( F0 + kx2)( 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
F
N=
− f cos θ + sin θ
10
Ni = = 13.49 lbf Ans.
−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= = = 0.967 Ans.
tan θ − f tan 60◦ − 0.25
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
=0= ∴ v* = 21.05 mph
dv 0.324
42.1(21.05) — 21.052
Q* = = 1367.6 cars/h Ans.
0.324
(b) v
l x l
2 2
v 0.324 −1
l
Q= = +
x +l v(42.1) − v2 v
Maximize Q with l = 10/5280 mi
v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434
1368 − 1221
% loss of throughput 12% Ans.
=
1221
SOLUTION MANUAL
, 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
cr F
D C cr
Facc
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)
F T
Fx = f N sin θ + N cos θ − = 0
r
T
r x F = N (sin θ − f cos θ ) Ans.
T = Nr ( f sin θ + cos θ )
N
fN
Combining
1 + f tan θ
T = Fr = KFr Ans. (2)
tan θ − 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 + kx2
From Prob. 1-6
f tan θ + 1
T2 = Fr
tan θ − f
( F0 + kx2)( 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
F
N=
− f cos θ + sin θ
10
Ni = = 13.49 lbf Ans.
−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= = = 0.967 Ans.
tan θ − f tan 60◦ − 0.25
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
=0= ∴ v* = 21.05 mph
dv 0.324
42.1(21.05) — 21.052
Q* = = 1367.6 cars/h Ans.
0.324
(b) v
l x l
2 2
v 0.324 −1
l
Q= = +
x +l v(42.1) − v2 v
Maximize Q with l = 10/5280 mi
v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434
1368 − 1221
% loss of throughput 12% Ans.
=
1221
