ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
TUTORIAL PROBLEM SET #6 (WEEK 44: PERIOD ENDING NOVEMBER 2, 2007)
The problem set provides a representative sample of questions on relevant course material and concepts
covered in the lectures. The tutorial problems sets are intended to develop good study habits and become
engaged in the learning process.
Tutorial Problem Set #6
Page 1 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
1: Problem 6-7 (page 279) Determine the force in each member of the truss and state if the members are
in tension or compression.
Tutorial Problem Set #6
Page 2 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
F 1 = 3 kN
F 2 = 8 kN
F 3 = 4 kN
F 4 = 10 kN
a = 2m
b = 1.5 m
θ = atan ⎛⎜
b⎞
⎟
⎝a⎠
Solution:
Initial Guesses
FBA = 1 kN
F BC = 1 kN
F AC = 1 kN
FAF = 1 kN
F CD = 1 kN
F CF = 1 kN
FDF = 1 kN
F ED = 1 kN
F EF = 1 kN
Given
Joint B
F 1 + FBC = 0
−F 2 − F BA = 0
Joint C
F CD − F BC − F AC cos ( θ ) = 0
−F 3 − F AC sin ( θ ) − F CF = 0
Joint E
Joint D
−F EF = 0
−F CD − F DF cos ( θ ) = 0
−F 4 − F DF sin ( θ ) − F ED = 0
Joint F
−F AF + F EF + FDF cos ( θ ) = 0
F CF + F DF sin ( θ ) = 0
⎛ FBA ⎞
⎜
⎟
⎜ FAF ⎟
⎜F ⎟
⎜ DF ⎟
⎜ FBC ⎟
⎜
⎟
⎜ FCD ⎟ = Find( FBA , FAF , FDF , FBC , FCD , FED , FAC , FCF , FEF )
⎜F ⎟
⎜ ED ⎟
⎜ FAC ⎟
⎜
⎟
⎜ FCF ⎟
⎟
⎜
⎝ FEF ⎠
⎛ FBA ⎞
⎜
⎟ ⎛ −8.00 ⎞
⎜ FAF ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 4.17 ⎟
DF
⎜
⎟ ⎜ 5.21 ⎟
⎜ FBC ⎟ ⎜ −3.00 ⎟
⎟
⎜
⎟ ⎜
⎜ FCD ⎟ = ⎜ −4.17 ⎟ kN
⎜ F ⎟ ⎜ −13.13 ⎟
⎟
⎜ ED ⎟ ⎜
⎜ FAC ⎟ ⎜ −1.46 ⎟
⎜
⎟ ⎜ −3.13 ⎟
⎟
⎜ FCF ⎟ ⎜
⎟ ⎝ 0.00 ⎠
⎜
F
⎝ EF ⎠
Positive means tension,
Negative means compression.
Tutorial Problem Set #6
Page 3 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
2: Problem 6-18 (page 281) Determine the force in each member of the truss and state if the members
are in tension or compression. Hint: The horizontal force component at A must be zero. Why?
Tutorial Problem Set #6
Page 4 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kip = 10 lb
Given:
F1 = 600 lb
F2 = 800 lb
a = 4 ft
b = 3 ft
θ = 60 deg
Solution:
Initial Guesses
FBA = 1 lb
FBD = 1 lb
FCB = 1 lb
FCD = 1 lb
Given
Joint C
Joint B
−FCB − F 2 cos ( θ ) = 0
FCB + F BD
b
2
a +b
−FCD − F2 sin ( θ ) = 0
=0
2
⎛ FBA ⎞
⎜
⎟
⎜ FBD ⎟
⎜
⎟ = Find( FBA , FBD , FCB , FCD)
F
CB
⎜
⎟
⎜F ⎟
⎝ CD ⎠
Tutorial Problem Set #6
−FBA − F BD
a
2
a +b
2
− F1 = 0
⎛ FBA ⎞
⎜
⎟ ⎛⎜ −1133.33 ⎞⎟
⎜ FBD ⎟ ⎜ 666.67 ⎟ Positive means Tension
lb Negative means
⎜
⎟=⎜
⎟ Compression
−
400.00
F
CB
⎜
⎟ ⎜
⎟
⎜ F ⎟ ⎝ −692.82 ⎠
⎝ CD ⎠
Page 5 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
3: Problem 6-28 (page 282) Determine the force in each member of the truss and state if the members
are in tension or compression. Set P1 = 2 kN, P2 = 4 kN.
Tutorial Problem Set #6
Page 6 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
P 1 = 2 kN
P 2 = 4 kN
a = 2m
θ = 15 deg
Solution:
Take advantage of the symmetry.
Initial Guesses:
F BD = 1 kN
F CD = 1 kN
F CA = 1 kN
F BC = 1 kN
Given
Joint D
Joint B
−P1
2
F AB = 1 kN
− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0
−P 2 cos ( 2 θ ) − FBC = 0
F BD − FAB − P2 sin ( 2θ ) = 0
Joint C
F CD cos ( θ ) − F CA cos ( θ ) = 0
(FCD + FCA)sin(θ) + FBC = 0
⎛⎜ FBD ⎞⎟
⎜ FCD ⎟
⎜
⎟
⎜ FAB ⎟ = Find(FBD , FCD , FAB , FCA , FBC )
⎜F ⎟
⎜ CA ⎟
⎜ FBC ⎟
⎝
⎠
⎛⎜ FBD ⎟⎞
⎛ −11.46 ⎞
⎜ FCD ⎟ ⎜ 6.69 ⎟
⎜
⎟ ⎜
⎟
⎜ FAB ⎟ = ⎜ −13.46 ⎟ kN
⎜ F ⎟ ⎜ 6.69 ⎟
⎜ CA ⎟ ⎜
⎟
⎜ FBC ⎟ ⎝ −3.46 ⎠
⎝
⎠
Tutorial Problem Set #6
⎛⎜ FFD
⎜ FED
⎜
⎜ FGF
⎜F
⎜ EG
⎜ FFE
⎝
⎛⎜ FFD
⎜ FED
⎜
⎜ FGF
⎜F
⎜ EG
⎜ FFE
⎝
⎟⎞ ⎛ −11.46 ⎞
⎟ ⎜ 6.69 ⎟
⎟ ⎜
⎟
⎟ = ⎜ −13.46 ⎟ kN
⎟ ⎜ 6.69 ⎟
⎟ ⎜
⎟
⎟ ⎝ −3.46 ⎠
⎠
F
⎟⎞ ⎛⎜ BD ⎞⎟
⎟ ⎜ FCD ⎟
⎟ ⎜
⎟
⎟ = ⎜ FAB ⎟
⎟ ⎜F ⎟
⎟ ⎜ CA ⎟
⎟ ⎜ FBC ⎟
⎠ ⎝
⎠
Positvive means Tension,
Negative means Compression
Page 7 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
4: Problem 6-33 (page 289) The roof truss supports the vertical loading shown. Determine the force in
members BC, CK, and KJ and state if these members are in tension or compression.
Tutorial Problem Set #6
Page 8 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
F1 = 4 kN
F2 = 8 kN
a = 2m
b = 3m
Solution:
Initial Guesses
A x = 1 kN
A y = 1 kN
FBC = 1 kN
FCK = 1 kN
FKJ = 1 kN
Given
Ax = 0
F2( 3a) + F1( 4a) − A y( 6a) = 0
⎛ 2b ⎞ + A ⎛ 2b ⎞ − A ( 2a) = 0
⎟
x⎜ 3 ⎟
y
⎝ 3 ⎠
⎝ ⎠
3a
⎞F = 0
⎛
FKJ + Ax + ⎜
BC
2
2⎟
⎝ b + 9a ⎠
FKJ ⎜
FCK + A y +
b
⎞F = 0
⎛
BC
⎜ 2
⎟
2
⎝ b + 9a ⎠
⎛⎜ Ax ⎟⎞
⎜ Ay ⎟
⎜
⎟
⎜ FKJ ⎟ = Find( Ax , Ay , FKJ , FCK , FBC )
⎜F ⎟
⎜ CK ⎟
⎜ FBC ⎟
⎝
⎠
Tutorial Problem Set #6
⎛⎜ Ax ⎟⎞
⎛ 0.00 ⎞
⎜ Ay ⎟ ⎜ 6.67 ⎟
⎜
⎟ ⎜
⎟
⎜ FKJ ⎟ = ⎜ 13.33 ⎟ kN Positive (T)
Negative (C)
⎜ F ⎟ ⎜ 0.00 ⎟
⎜ CK ⎟ ⎜
⎟
−
14.91
⎝
⎠
⎜ FBC ⎟
⎝
⎠
Page 9 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
5: Problem 6-39 (page 290) Determine the force members BC, FC, and FE, and state if the members are
in tension or compression.
Tutorial Problem Set #6
Page 10 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 6 kN
a = 3m
b = 3m
Solution:
θ = atan ⎛⎜
a⎞
⎟
⎝ b⎠
Initial Guesses
Dy = 1 kN
F BC = 1 kN
F FC = 1 kN
F FE = 1 kN
Given
−F 1 b − F 2( 2b) + Dy( 3b) = 0
Dy b − FFE cos ( θ ) a = 0
(
)
−F FC − F BC + FFE cos ( θ ) = 0
(
)
−F 2 + Dy + F FE + FBC sin ( θ ) = 0
⎛ Dy
⎜
⎜ FBC
⎜
⎜ FFC
⎜F
⎝ FE
⎞
⎟
⎟
⎟ = Find( Dy , FBC , FFC , FFE )
⎟
⎟
⎠
⎛ Dy
⎜
⎜ FBC
⎜
⎜ FFC
⎜F
⎝ FE
⎞
⎟ ⎛⎜ 6 ⎟⎞
⎟ ⎜ −8.49 ⎟
kN
⎟=⎜
⎟
0
⎟ ⎜
⎟
⎟ ⎝ 8.49 ⎠
⎠
Tutorial Problem Set #6
Positive (T)
Negative (C)
Page 11 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
6: Problem 6-51 (page 292) Determine the force developed in members BC and CH of the roof truss and
state if the members are in tension or compression.
Tutorial Problem Set #6
Page 12 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
F 1 = 1.5 kN
F 2 = 2 kN
a = 1.5 m
b = 1m
c = 2m
d = 0.8 m
Solution:
θ = atan ⎛⎜
a⎞
⎟
⎝c⎠
φ = atan ⎛⎜
⎞
⎟
⎝c−b⎠
a
Initial Guesses:
E y = 1 kN
F BC = 1 kN
F CH = 1 kN
Given
−F 2( d) − F 1( c) + E y( 2c) = 0
F BC sin ( θ ) ( c) + F CH sin ( φ ) ( c − b) + E y( c) = 0
−F BC sin ( θ ) − F CH sin ( φ ) − F 1 + E y = 0
⎛ Ey
⎜
⎜ F BC
⎜F
⎝ CH
⎞
⎟
⎟ = Find( Ey , F BC , FCH ) Ey = 1.15 kN
⎟
⎠
Tutorial Problem Set #6
⎛⎜ FBC
⎜ FCH
⎝
⎟⎞ = ⎛ −3.25 ⎞ kN Positive (T)
Negative (C)
⎟ ⎜⎝ 1.923 ⎟⎠
⎠
Page 13 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
7: Problem 6-70 (page 314) The 150-lb man attempts to lift himself and the 10-lb seat using the rope and
pulley system shown. Determine the force at A needed to do so, and also find his reaction on the seat.
Tutorial Problem Set #6
Page 14 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Given:
W1 = 150 lb
W2 = 10 lb
Solution:
Pulley C:
ΣFy = 0;
3T − R = 0
Pulley B:
ΣFy = 0;
3R − P = 0
Thus,
P = 9T
Man and seat:
ΣFy = 0;
T + P − W1 − W2 = 0
10T = W1 + W2
T =
W1 + W2
10
P = 9T
T = 16 lb
P = 144 lb
Seat:
ΣFy = 0;
P − N − W2 = 0
N = P − W2
Tutorial Problem Set #6
N = 134 lb
Page 15 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
8: Problem 6-76 (page 315) The compound beam is fixed supported at A and supported by rockers at B
and C. If there are hinges (pins) at D and E, determine the reactions at the supports A, B, and C.
Tutorial Problem Set #6
Page 16 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
a = 2m
M = 48 kN ⋅ m
b = 4m
w1 = 8
kN
w2 = 6
kN
c = 2m
d = 6m
m
m
e = 3m
Solution:
Guesses
Ax = 1 N
Ay = 1 N
MA = 1 N m
Dx = 1 N
Dy = 1 N
By = 1 N
Ey = 1 N
Ex = 1 N
Cy = 1 N
Given
A y − w2 a − Dy = 0
M A − w2 a
−w1
( b + c)
−w1 ⎛⎜
⎝
d+e
2
Dy − w1 ( b + c) + B y − E y = 0
2
2
E y − w1
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
a
− Dy a = 0
2
− A x − Dx = 0
+ B y b − E y( b + c) = 0
+ Cy = 0
Dx + E x = 0
−E x = 0
d + e ⎞⎛ d + e ⎞
⎟⎜
⎟ + Cy d − M = 0
2 ⎠⎝ 3 ⎠
⎞
⎟
Ay
⎟
MA ⎟
⎟
Dx ⎟
Dy ⎟ = Find( A x , A y , MA , Dx , Dy , B y , E y , E x , Cy)
⎟
By ⎟
⎟
Ey
⎟
Ex ⎟
⎟
Cy ⎠
Ax
Tutorial Problem Set #6
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ = ⎜ ⎟ kN
⎝ Ay ⎠ ⎝ 19 ⎠
MA = 26 kN m
B y = 51 kN
Cy = 26 kN
Page 17 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
9: Problem 6-83 (page 316) The wall crane supports a load of 700 lb. Determine the horizontal and
vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W?
Tutorial Problem Set #6
Page 18 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
3
kip = 10 lb
Units Used:
Given:
F = 700 lb
a = 4 ft
b = 4 ft
c = 4 ft
θ = 60 deg
Solution:
Pulley E:
+
↑Σ Fy = 0;
T =
2T − F = 0
1
F
2
T = 350 lb
This is the force in the cable at the winch W
a⎞
⎟
⎝b⎠
φ = atan ⎛⎜
Member ABC:
Σ MA = 0;
−F ( b + c) + ( TBD sin ( φ ) − T sin ( θ ) ) b = 0
F ⎛⎜
TBD =
⎝
b+ c⎞
⎟ + T sin ( θ )
b ⎠
sin ( φ )
TBD = 2408.56 lb
+
↑Σ Fy = 0;
− A y + TBD sin ( φ ) − T sin ( θ ) − F = 0
A y = TBD sin ( φ ) − T sin ( θ ) − F
A y = 700.00 lb
+
Σ F x = 0;
→
A x − TBD cos ( φ ) − T cos ( θ ) = 0
A x = TBD cos ( φ ) + T cos ( θ )
A x = 1.878 kip
At D:
Dx = TBD cos ( φ )
Dx = 1.703 kip
Dy = TBD sin ( φ )
Dy = 1.703 kip
Tutorial Problem Set #6
Page 19 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
10: Problem 6-102 (page 321) The pillar crane is subjected to the load having a mass of 500 kg.
Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin
support C when the boom is tied in the position shown.
Tutorial Problem Set #6
Page 20 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Units Used:
3
kN = 10 N
Given:
M = 500 kg
a = 1.8 m
b = 2.4 m
θ 1 = 10 deg
θ 2 = 20 deg
g = 9.81
m
s
2
Solution:
initial guesses: FCB = 10 kN
FAB = 10 kN
Given
−M
2
−M
2
g cos ( θ 1 ) − FAB cos ( θ 2 ) + F CB
b
=0
2
a +b
g sin ( θ 1 ) − FAB sin ( θ 2 ) + F CB
⎛ FAB ⎞
⎜
⎟ = Find( FAB , FCB)
F
CB
⎝
⎠
Tutorial Problem Set #6
a
2
a +b
⎛ Cx ⎞
⎜ ⎟ =
⎝ Cy ⎠
2
− Mg = 0
2
⎛ ⎞
⎜ ⎟
2
2 a
a +b ⎝ ⎠
F CB
b
⎛ FAB ⎞ ⎛ 9.7 ⎞
⎜
⎟ ⎜
⎟
⎜ Cx ⎟ = ⎜ 11.53 ⎟ kN
⎜ C ⎟ ⎝ 8.65 ⎠
⎝ y ⎠
Page 21 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
11: Problem 6-107 (page 322) A 5-lb force is applied to the handles of the vise grip. Determine the
compressive force developed on the smooth bolt shank A at the jaws.
Tutorial Problem Set #6
Page 22 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Given:
F = 5 lb
b = 1 in
a = 1.5 in
c = 3 in
d = 0.75 in
e = 1 in
θ = 20 deg
Solution:
From FBD (a)
ΣME = 0;
d+e
⎤b = 0
⎢ 2
2⎥
⎣ c + ( d + e) ⎦
F( b + c) − FCD⎡
⎡ c2 + ( d + e) 2 ⎤
⎥
⎣ b( d + e) ⎦
FCD = F ( b + c) ⎢
ΣFx = 0;
FCD = 39.693 lb
c
⎤
⎢ 2
2⎥
⎣ c + ( d + e) ⎦
Ex = F CD⎡
Ex = 34.286 lb
From FBD (b)
ΣMB = 0;
NA sin ( θ ) d + NA cos ( θ ) a − Ex( d + e) = 0
NA = Ex
d+e
⎞
⎛
⎜ ( )
⎟
⎝ sin θ d + cos ( θ ) a ⎠
NA = 36.0 lb
Tutorial Problem Set #6
Page 23 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
12: Problem 6-125 (page 327) The four-member “A” frame is supported at A and E by smooth collars and
at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there
is 800 N, determine the largest vertical force P that can be supported by the frame. Also, what are the x,
y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and
the pin at G only exert force components on the frame.
Tutorial Problem Set #6
Page 24 of 25
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
Given:
Fmax = 800 N
a = 300 mm
b = 600 mm
c = 600 mm
Solution:
ΣMx = 0;
b
−P 2 c +
2
2
F max c = 0
b +c
F max b
P =
2
2
P = 282.84 N
2
b +c
c
Bz + Dz − F max
2
=0
2
Dz = B z
b +c
Fmax c
Bz =
2
2
Dz = B z
2
Bz = 283 N
b +c
Dz = 283 N
b
By + Dy − F max
2
=0
2
Dy = By
b +c
F max b
By =
2
2
2
Dy = By
By = 283 N
b +c
Dy = 283 N
Bx = Dx = 0
Tutorial Problem Set #6
Page 25 of 25