ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
TUTORIAL PROBLEM SET #4 SOLUTION (WEEK 40/41: PERIOD ENDING OCTOBER 13, 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. The mid-term exam on October 18 will cover topics up to and including
Tutorial Problem Set #4.
1: Problem 4-6 (page 133) Determine the magnitude of the force F that should be applied at the end of
the lever such that this force creates a clockwise moment M about point O.
Given:
M = 15 N m
θ = 30 deg
φ = 60 deg
a = 50 mm
b = 300 mm
Solution:
M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )
F =
M
cos ( θ ) ( a + b sin ( φ ) ) − sin ( θ ) ( b cos ( φ ) )
F = 77.6 N
Tutorial Problem Set #4
Page 1 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
2: Problem 4-10 (page 134) A force F is applied to the wrench. Determine the moment of this force about
point O. Solve the problem using both a scalar analysis and a vector analysis.
Given:
F = 40 N
θ = 20 deg
a = 30 mm
b = 200 mm
Scalar Solution
MO = −F cos ( θ ) b + F sin ( θ ) a
MO = −7.11 N ⋅ m
MO = 7.11 N ⋅ m
Vector Solution
⎛ b ⎞ ⎛ −F sin ( θ ) ⎞
⎜ ⎟ ⎜
⎟
MO = a × −F cos ( θ )
⎜ ⎟ ⎜
⎟
0
⎝0⎠ ⎝
⎠
Tutorial Problem Set #4
⎛ 0 ⎞
⎜ 0 ⎟ N ⋅m
MO =
⎜
⎟
⎝ −7.11 ⎠
MO = 7.107 N ⋅ m
Page 2 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
3: Problem 4-21 (page 136) The tool at A is used to hold a power lawnmower blade stationary while the
nut is being loosened with the wrench. If a force P is applied to the wrench at B in the direction shown,
determine the moment it creates about the nut at C. What is the magnitude of force F at A so that it
creates the opposite moment about C?
Given:
P = 50 N
θ = 60 deg
a = 400 mm
c = 5
b = 300 mm
d = 12
Solution:
(a)
(b)
MA = P sin ( θ ) b
MA − F
MA = 13.0 N ⋅ m
d
2
a=0
2
c +d
⎛ c2 + d2 ⎞
⎟
F = MA ⎜
⎝ da ⎠
Tutorial Problem Set #4
F = 35.2 N
Page 3 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
4: Problem 4-39 (page 139) The curved rod lies in the x-y plane and has a radius r. If a force F acts at its
end as shown, determine the moment of this force about point B.
Given:
F = 80 N
a = 1m
b = 2m
c = 3m
r= 3m
θ = 45 deg
Solution:
rAC
⎛ a ⎞
⎜ ⎟
= −c
⎜ ⎟
⎝ −b ⎠
MB = rBA × Fv
Tutorial Problem Set #4
Fv = F
rAC
rAC
rBA
⎛ r cos ( θ ) ⎞
⎜
⎟
= r − r sin ( θ )
⎜
⎟
0
⎝
⎠
⎛ −37.6 ⎞
⎜ 90.7 ⎟ N ⋅m
MB =
⎜
⎟
⎝ −154.9 ⎠
Page 4 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
5: Problem 4-45 (page 140) The pipe assembly is subjected to the force F. Determine the moment of this
force about point B.
Given:
F = 80 N
a = 400 mm
b = 300 mm
c = 200 mm
d = 250 mm
θ = 40 deg
φ = 30 deg
Solution:
rBC
⎛b+d⎞
⎜ 0 ⎟
=
⎜
⎟
⎝ −c ⎠
⎛ 550.00 ⎞
⎜ 0.00 ⎟ mm
rBC =
⎜
⎟
⎝ −200.00 ⎠
⎛⎜ cos ( φ ) sin ( θ ) ⎟⎞
Fv = F ⎜ cos ( φ ) cos ( θ ) ⎟
⎜ −sin ( φ ) ⎟
⎝
⎠
MB = rBC × Fv
Tutorial Problem Set #4
⎛ 44.53 ⎞
⎜
⎟
Fv = 53.07 N
⎜
⎟
⎝ −40.00 ⎠
⎛ 10.61 ⎞
⎜
⎟
MB = 13.09 N ⋅ m
⎜
⎟
⎝ 29.19 ⎠
Page 5 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
6: Problem 4-54 (page 148) The force F is applied to the handle of the box wrench. Determine the
component of the moment of this force about the z axis which is effective in loosening the bolt
Given:
a = 3 in
b = 8 in
c = 2 in
⎛ 8 ⎞
F = ⎜ −1 ⎟ lb
⎜ ⎟
⎝ 1 ⎠
Solution:
⎛0⎞
⎜ ⎟
k = 0
⎜ ⎟
⎝1⎠
⎛ c ⎞
⎜ ⎟
r = −b
⎜ ⎟
⎝ a ⎠
Mz = ( r × F) ⋅ k
Mz = 62 lb⋅ in
7: Problem 4-57 (page 149) The cutting tool on the lathe exerts a force F on the shaft in the direction
shown. Determine the moment of this force about the y axis of the shaft.
Units Used:
3
kN = 10 N
Given:
⎛ 6 ⎞
⎜ ⎟
F = −4 kN
⎜ ⎟
⎝ −7 ⎠
a = 30 mm
θ = 40 deg
Solution:
⎛ cos ( θ ) ⎞
⎜ 0 ⎟
r = a
⎜
⎟
(
)
sin
θ
⎝
⎠
Tutorial Problem Set #4
⎛0⎞
⎜ ⎟
j= 1
⎜ ⎟
⎝0⎠
My = ( r × F) ⋅ j
My = 0.277 kN ⋅ m
Page 6 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
8: Problem 4-63 (page 150) Determine the magnitude of the moment of the force F about the base line
CA of the tripod.
Given:
⎛ 50 ⎞
F = ⎜ −20 ⎟ N
⎜
⎟
⎝ −80 ⎠
a = 4m
b = 2.5 m
c = 1m
d = 0.5 m
e = 2m
f = 1.5 m
g = 2m
Solution:
rCA
⎛ −g ⎞
⎜ e ⎟ u = rCA
=
⎜ ⎟ CA
rCA
⎝ 0 ⎠
Tutorial Problem Set #4
rCD
⎛b−g⎞
⎜ e ⎟ M = ( r × F) ⋅u
=
CA
CD
CA
⎜
⎟
⎝ a ⎠
MCA = 226 N ⋅ m
Page 7 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
9: Problem 4-75 (page 160) Three couple moments act on the pipe assembly. Determine the magnitude
of M3 and the bend angle θ so that the resultant couple moment is zero.
Given:
θ 1 = 45 deg
M1 = 900 N ⋅ m
M2 = 500 N ⋅ m
Solution:
Initial guesses:
θ = 10 deg
M3 = 10 N ⋅ m
Given
+
ΣMx = 0;
→
M1 − M3 cos ( θ ) − M2 cos ( θ 1 ) = 0
+
M3 sin ( θ ) − M2 sin ( θ 1 ) = 0
↑ ΣMy = 0;
⎛ θ ⎞
⎜
⎟ = Find( θ , M3 )
⎝ M3 ⎠
Tutorial Problem Set #4
θ = 32.9 deg
M3 = 651 N ⋅ m
Page 8 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
10: Problem 4-93 (page 164) Express the moment of the couple acting on the rod in Cartesian vector
form. What is the magnitude of the couple moment?
Given:
⎛ 14 ⎞
⎜ ⎟
F = −8 N
⎜ ⎟
⎝ −6 ⎠
a = 1.5 m
b = 0.5 m
c = 0.5 m
d = 0.8 m
Solution:
⎛ d ⎞
⎛0⎞
⎜
⎟
⎜ ⎟
M = a × F + 0 × ( −F)
⎜ ⎟
⎜ ⎟
⎝ −c ⎠
⎝b⎠
Tutorial Problem Set #4
⎛ −17 ⎞
⎜
⎟
M = −9.2 N ⋅ m
⎜
⎟
⎝ −27.4 ⎠
M = 33.532 N ⋅ m
Page 9 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
11: Problem 4-111 (page 180) The forces and couple moments which are exerted on the toe and heel
plates of a snow ski are Ft, Mt, and Fh, Mh, respectively. Replace this system by an equivalent force and
couple moment acting at point P. Express the results in Cartesian vector form.
Given:
a = 120 mm
b = 800 mm
⎛ −50 ⎞
⎜ 80 ⎟ N
Ft =
⎜
⎟
⎝ −158 ⎠
⎛ −6 ⎞
⎜ 4 ⎟ N ⋅m
Mt =
⎜ ⎟
⎝ 2 ⎠
⎛ −20 ⎞
⎜ 60 ⎟ N
Fh =
⎜
⎟
⎝ −250 ⎠
⎛ −20 ⎞
⎜ 8 ⎟ N ⋅m
Mh =
⎜
⎟
⎝ 3 ⎠
Solution:
FR = Ft + Fh
⎛ −70 ⎞
⎜
⎟
FR = 140 N
⎜
⎟
⎝ −408 ⎠
⎛b⎞
⎛a+b⎞
⎜ ⎟
⎜ 0 ⎟×F
MP = Mt + Mh + 0 × Fh +
⎜ ⎟
⎜
⎟ t
0
0
⎝ ⎠
⎝
⎠
⎛ −26 ⎞
⎜
⎟
MP = 357.4 N ⋅ m
⎜
⎟
⎝ 126.6 ⎠
Tutorial Problem Set #4
Page 10 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
12: Problem 4-121 (page 182) Replace the loading on the frame by a single resultant force. Specify
where its line of action intersects member CD, measured from end C.
Given:
F1 = 500 N
a = 3m
b = 2m
F2 = 300 N
c = 1m
F3 = 250 N
d = 2m
M = 400 N ⋅ m
e = 3m
θ = 60 deg
f = 3
g = 4
Solution:
⎛
⎞ − F ( cos ( θ ) )
1
⎝ g +f ⎠
g
FRx = −F 3 ⎜
⎛
FRy = −F 2 − F 3 ⎜
⎝
FR =
2⎟
2
⎞ − F sin ( θ )
1
f +g ⎠
2
f
2
FRx + F Ry
2⎟
2
FRx = −450 N
FRy = −883.0127N
FR = 991 N
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
θ 1 = 63 deg
⎛
⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ )
1
1
⎟
2
2
+
f
g
⎝
⎠
FRy( x) = M − F2 ( d + c) − F 3 ⎜
⎛
⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ )
1
1
⎟
2
2
+
f
g
⎝
⎠
M − F 2 ( d + c) − F3 ⎜
x =
f
f
FRy
x = 2.64m
Tutorial Problem Set #4
Page 11 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
13: Problem 4-131 (page 183) Handle forces F1 and F2 are applied to the electric drill. Replace this
system by an equivalent resultant force and couple moment acting at point O. Express the results in
Cartesian vector form.
Given:
a = 0.15 m
b = 0.25 m
c = 0.3 m
⎛ 6 ⎞
F1 = ⎜ −3 ⎟ N
⎜
⎟
⎝ −10 ⎠
⎛ 0 ⎞
F2 = ⎜ 2 ⎟ N
⎜ ⎟
⎝ −4 ⎠
Solution:
FR = F1 + F2
⎛ 6 ⎞
⎜
⎟
FR = −1 N
⎜
⎟
⎝ −14 ⎠
⎛a⎞
⎛ 0 ⎞
⎜
⎟
⎜ ⎟
MO = 0 × F1 + −b × F2
⎜ ⎟
⎜ ⎟
⎝c⎠
⎝ c ⎠
⎛ 1.3 ⎞
⎜ 3.3 ⎟ N ⋅m
MO =
⎜
⎟
⎝ −0.45 ⎠
Tutorial Problem Set #4
Page 12 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
14: Problem 4-142 (page 190) Replace the loading by a single resultant force, and specify the location of
the force on the beam measured from point O.
Units Used:
3
kN = 10 N
Given:
w = 6
kN
m
F = 15 kN
M = 500 kN ⋅ m
a = 7.5 m
b = 4.5 m
Solution:
Initial Guesses:
FR = 1 kN
d = 1m
Given
FR =
1
w( a + b) + F
2
−FR d = − M −
⎛ 1 w a ⎞ ⎛ 2 a ⎞ − ⎛ 1 w b ⎞ ⎛ a + b ⎞ − F( a + b
⎜
⎟⎜ ⎟ ⎜
⎟⎜
⎟
3⎠
⎝2
⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝
⎛ FR ⎞
⎜
⎟ = Find( FR , d)
d
⎝
⎠
Tutorial Problem Set #4
FR = 51 kN
d = 17.922 m
Page 13 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
15: Problem 4-149 (page 192) The distribution of soil loading on the bottom of a building slab is shown.
Replace this loading by an equivalent resultant force and specify its location, measured from point O.
Units Used:
3
kip = 10 lb
Given:
w1 = 50
lb
ft
w2 = 300
lb
ft
w3 = 100
lb
ft
a = 12 ft
b = 9 ft
Solution:
(w2 − w1)a + 2 (w2 − w3)b + w3 b
2
FR = w1 a +
1
FR d = w1 a
a
d =
2
1
+
(w2 − w1)a
2
2a
2
2
1
3
+
FR = 3.9 kip
b
b
w2 − w3 ) b⎛⎜ a + ⎟⎞ + w3 b⎛⎜ a + ⎟⎞
(
2
3
2
1
⎝
2
⎠
⎝
2
3 w3 b a + 2 w3 b + w1 a + 2 a w2 + 3 b w2 a + w2 b
Tutorial Problem Set #4
6FR
⎠
d = 11.3 ft
Page 14 of 15
ENGI 1313 Mechanics I
Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng.
Fall 2007
Revision 0
16: Problem 4-155 (page 193) Determine the equivalent resultant force and couple moment at point O.
Units Used:
3
kN = 10 N
Given:
a = 3m
wO = 3
kN
m
⎛x⎞
⎟
⎝a⎠
2
w ( x) = wO ⎜
Solution:
⌠a
FR = ⎮ w ( x) dx
⌡0
FR = 3 kN
⌠a
MO = ⎮ w ( x) ( a − x) dx
⌡0
MO = 2.25 kN ⋅ m
Tutorial Problem Set #4
Page 15 of 15