NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
PROBLEM 8-39
Two blocks A and B have a weight of
10 lb and 6 lb, respectively. They are
resting on the incline for which the
coefficients of static friction are
μ A = 0.15 and μ B = 0.25 . Determine
the angle θ which will cause motion
of one of the blocks. What is the
friction force under each of the blocks
when this occurs? The spring has a
stiffness of k = 2lb / ft and is
originally unstreched.
SOLUTION 8-39:
(a)
(b)
Equations of Equilibrium: Since block A and B is either not moving or on the verge of
moving, the spring force Fsp = 0 ,
FBD (a):
∑F
∑F
x
=0
FA − 10 sin θ = 0
(1)
y
=0
N A − 10 cos θ = 0
(2)
From FBD (b):
∑F
∑F
x
=0
FB − 6 sin θ = 0
(3)
y
=0
N B − 6 cos θ = 0
(4)
Friction: Assuming block A is on the verge of slipping, then:
PAGE
1
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
FA = μ sA N A = 0.15 N A
Solving Equations (1), (2), (3), (4) and (5) yields:
θ = 8.531D , FA = 1.483lb , N A = 9.889lb FB = 0.8900lb , N B = 5.934lb
Since ( FB ) max = μ sB N B = 0.25(5.934) = 1.483lb > FB , block B doesn’t slip. Therefore the
above assumption is correct, thus:
θ = 8.531D ,
FA = 1.483lb ,
FB = 0.8900lb
PROBLEM 8-46
Each of the cylinders has a mass of 50
kg. If the coefficient of static friction at
the points of contact are
μ A = μ B = μ C = 0.5 and μ D = 0.6 ,
determine the couple moment M
needed to rotate cylinder E.
SOLUTION 8-46:
(a)
(b)
PAGE
2
14
NAME & ID
DATE
PAGE
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
Mechatronics Engineering
3
FBD (a):
∑F
∑F
∑M
x
=0
N D − FC = 0
(1)
y
=0
N C + FD − 490.5 = 0
(2)
M − FC (0.3) − FD (0.3) = 0
(3)
O
=0
FBDD (b):
∑F
∑F
∑M
x
=0
N A − FB − N D = 0
(4)
y
=0
N B + FA − FD − 490.5 = 0
(5)
FA (0.3) − FB (0.3) − FD (0.3) = 0
(6)
P
=0
Friction: Assuming cylinder E slips at points C and D and cylinder F does not move, then:
FC = μ sC N C = 0.5 N C
and FD = μ sD N D = 0.6 N D .
Equations (1) and (2) and (3) and solving, we have:
N C = 377.31N
N D = 188.65 N
Substituting
M = 90.6 N .m
these
values
into
Ans
If cylinder F is on the verge of slipping at point A, then FA = μ sA N A = 0.5 N A . Substitute
this value into Equations (4), (5) and (6) and solving, we have:
N A = 150.92 N
N B = 679.15 N
FB = 37.73 N
Ans
Since ( FB ) max = μ sB N B = 0.5(679.15) = 339.58 N > FB , cylinder F does not move.
Therefore the above assumption is correct.
EXTRA PRACTICE PROBLEM 8- 63
Determine the largest weight of the wedge
that can be placed between the 8-lb cylinder
and the wall without upsetting equilibrium.
The coefficient of static friction at A and C is
μ S = 0.5 and at B, μ ' S = 0.6 . Determine
the couple moment M needed to rotate
cylinder E.
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
Mechatronics Engineering
EXTRA PRACTICE SOLUTION 8-63:
(a)
(b)
Equations of equilibrium:
FBD (a):
∑F
∑F
x
=0
N B cos 30 D − FB cos 60 D − N C = 0
(1)
y
=0
N B sin 30 D + FB sin 60 D + FC − W = 0
(2)
FBD (b):
∑F
∑F
∑M
x
=0
N A − N B sin 30 D − FB sin 60 D − 8 = 0
(3)
y
=0
FA + FB cos 60 D − N B cos 30 D = 0
(4)
FA (0.5) − FB (0.5) = 0
(5)
O
=0
Friction: Assume slipping occurs at points C and A, then FC = μ sC N C = 0.5 N C
and FA = μ sA N A = 0.5 N A . Substituting these values into Equations (1),(2), (3),(4) and (5)
and solving, we have:
W = 66.64lb
N B = 51.71lb
N A = 59.71lb
Ans
FB = N C = 29.86lb
PAGE
4
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
PROBLEM 8-70
If the beam AD is loaded as
shown, determine the horizontal
force P which must be applied to
the wedge in order to remove it
from under the beam. The
coefficient of static friction at the
wedge’s top and bottom surfaces
are μ CA = 0.25 and μ CB = 0.35
respectively. If P=0, is the wedge
self locking? Neglect the weight
and size of the wedge and the
thickness of the beam.
SOLUTION 8-70:
(a)
PAGE
5
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
(b)
Equations of equilibrium and friction:
If the wedge is on the verge of moving to the right, then slipping will have to occur at both
contact surfaces. Thus, FA = μ sA N A = 0.25 N A and FB = μ sB N B = 0.35 N B .
FBD (a):
∑M
D
=0
N A cos10 D (7) + 0.25 N A sin 10 D (7) − 6(2) − 16(5) = 0
N A = 12.78kN
FBD (b):
∑F
y
=0
N B − 12.78 sin 80 D − 0.25(12.78) sin 10 D = 0
N B = 13.14kN
∑F
x
=0
P + 12.78 cos 80 D − 0.25(12.78) cos10 D − 0.35(13.14) = 0
P = 5.53kN
Ans
Since a force P(>0) is required to pull out the wedge, the wedge will be self-locking when
P = 0. Ans
EXTRA PRACTICE PROBLEM 9-13
The plate has a thickness of 0.25 ft and a
specific weight of γ = 180 lb/ft 3 . Determine
the location of the center of gravity. Also,
find the tension in each of the cords used to
support it.
PAGE
6
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
Mechatronics Engineering
EXTRA PRACTICE SOLUTION 9-13:
y
Calculation of the area and the
first moments:
Height of the differential element:
(x,y)
1
y = x − 8 x 2 + 16
16
( x , y )
Area of the differential element:
1
⎛
⎞
dA = ydx = ⎜ x − 8 x 2 + 16 ⎟ dx
⎝
⎠
x
dx
16
Centroid of differential element:
1
⎞
1⎛
x = x and y = ⎜ x − 8 x 2 + 16 ⎟
2⎝
⎠
The total area is given by,
16
1
⎛
⎞
⎛ x 2 16 3
⎞
A = ∫ dA = ∫ ⎜ x − 8 x 2 + 16 ⎟dx = ⎜ − x 2 + 16 x ⎟ = 42.67 ft 2
⎠
⎝ 2 3
⎠0
A
0 ⎝
16
Therefore:
16
1
⎡⎛
⎞ ⎤ ⎛ x 3 16 52
⎞
2
= ∫ x ⎢⎜ x − 8 x + 16 ⎟ dx ⎥ = ⎜ − x + 8 x 2 ⎟ = 136.53 ft 3
∫A xdA
⎠ ⎦⎥ ⎝ 3 5
⎠0
0
⎣⎢⎝
16
16
1
1
⎞ ⎡⎛
⎞ ⎤
1⎛
2
2
8
16
8
16
ydA
=
x
−
x
+
x
−
x
+
⎢
⎜
⎟
⎜
⎟ dx ⎥
∫A
∫0 2 ⎝
⎢
⎠ ⎣⎝
⎠ ⎥⎦
16
⎞
1 ⎛ 1 3 32 52
512 32
2
3
ydA
=
x
−
x
+
x
−
x
+
x
48
256
⎜
⎟ = 136.53 ft
∫A
2⎝3
5
3
⎠0
The coordinates of the centroid are therefore given by:
PAGE
7
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
Mechatronics Engineering
x =
∫ xdx
A
∫ dA
=
136.53
= 3.2 ft
42.67
ANS
=
136.53
= 3.2 ft
42.67
ANS
A
y =
∫ ydx
A
∫ dA
A
Tension in the cables:
The weight of the plate is: W= 42.67(0.25)(180)=1920 lb
TB
W
TA
TC
Equations of equilibrium:
∑M
∑M
x
= 0; 1920(3.20) − TA (16) = 0
TA = 384 lb
ANS
y
= 0; TC (16 ) − 1920(3.2) = 0
TC = 384 lb
ANS
TB = 1152 lb
ANS
PROBLEM 9- 42
The hemisphere of radius r is made
from a stack of very thin plates
such that the density varies with
height = kz, where k is a constant,
Determine its mass and the
−
distance z to the center of mass
G.
PAGE
8
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
SOLUTION 9-42:
Mass and Moment Arm: The density of the material is ρ = kz . The mass of the thin disk
−
differential element is dm = ρdV = ρπy 2 dz = kz (π (r 2 − z 2 )dz ) and its Centroid z = z .
Evaluating the integrals, we have:
r
m = ∫ dm = ∫ kz (π (r 2 − z 2 )dz )
0
m
r 2 z 2 z 2 r πkr 4
= πk (
− ) |0 =
2
4
4
−
Ans
r
2
2
∫ z dm = ∫ z{kz[(π (r − z )dz ]}
0
m
= πk (
r 2 z 2 z 2 r 2πkr 5
− )) |0 =
3
5
15
Centroid: Applying Equation 9-4, we have:
−
∫ z dm
m
∫ dm
m
=
2πkr 5 / 15 8
= r
πkr 4 / 4 15
PAGE
9
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
EXTRA PRACTICE PROBLEM 9-51
The three members of the frame
each have a weight per unite length
of 4 lb/ft. Locate the position
−
−
( x, y ) of the center of gravity.
Neglect the size of the pins at the
joints and the thickness of the
members. Also, calculate the
reactions at the fixed support A.
SOLUTION 9-51:
PAGE
10
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
−
∑ x.W = 1.5(4) 45 + 3(4) 72 = 142.073lb. ft
∑ W = 4(7) + 4 45 + 4 72 = 88.774lb
−
∑ x.W
x=
∑W
−
=
142.073
= 1.06 ft Ans
88.774
−
∑ y.W = 3.5(4)(7) + 7(4)
45 + 10(4) 72 = 625.241lb. ft
−
∑ y.W
y=
∑W
−
∑F
∑F
∑M
=
625.241
= 7.04 ft Ans
88.774
x
=0
Ax = 0
y
=0
Ay = 88.774 + 60 = 149lb
A
=0
− 60(6) − 88.774(1.06) + M A = 0
M A = 502lb. ft
Ans
PROBLEM 9- 67
−
Locate the Centroid y of the
beam’s cross-section built up from
a channel and a wide-flange beam.
PAGE
11
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
SOLUTION 9-67:
Centroid: The area of each segment and its respective centroid are tabulated below:
Segment A(in 2 )
−
−
y (in)
y
A ( in
1
14(0.4)
16.20
90.72
2
3.4(1.3)
14.7
64.97
3
10.3(0.76)
15.62
122.27
4
14.48(0.56)
8.00
64.87
5
10.3(0.76)
0.38
2.97
∑
33.78
345.81
Thus:
−
∑ y. A = 345.81 = 10.24in = 10.2in
y=
∑ A 33.78
−
Ans
)
PAGE
12
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
EXTRA PRACTICE PROBLEM 9-115
The storage tank contains oil
having a specific weight
of γ 0 = 56lb / ft 3 . If the tank is 6
ft wide, calculate the resultant
force acting on the inclined side
BC of the tank, caused by the oil,
and specify its location along BC,
measured from B. Also compute
the total resultant force acting on
the bottom of the tank.
EXTRA PRACTICE SOLUTION 9-115:
WB = bγ 0 h = 6(56)(2) = 672lb / ft
WC = bγ 0 h = 6(56)(10) = 3360lb / ft
Fh1 = 8(672) = 5376lb
Fh 2 = 1 / 2(3360 − 672)(8) = 10752lb
FV 1 = (3)(2)(6)(56) = 2016lb
FV 2 = 1 / 2(3)(8)(6)(56) = 4032lb
∑F
∑F
x
=0
FRx = 5376 + 10752 = 16128lb
y
=0
FRy = 2016 + 4032 = 6048lb
FR = (16128) 2 + (6048) 2 = 17225lb = 17.2kip Ans
6048
) = 20.56 D
16128
θ = tan −1 (
PAGE
13
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 9
SOLUTIONS
Mechatronics Engineering
∑M
RB
17225d = 10752(2 / 3)(8) + 5376(4) + 2016(1.5) + 4032(2)
=0
d = 5.22 ft
Ans
At bottom:
FR = 4(14)(6)(56) = 18816lb = 18.8kip
Problem 9-119
The pressure loading on the
plate is described by the
function:
p = 10[6 /( x + 1) + 8]lb / ft 2
Determine the magnitude of
the resultant force and the
−
−
coordinates ( x, y ) of the point
where the line of action of the
force intersects the plate.
SOLUTION 9-119:
P = 10[
6
+ 8]
( x + 1)
2
FR = ∫ PdA = ∫ 10[
0
6
+ 8].3dx
( x + 1)
FR = 30(6 ln( x + 1) + 8 x |02 = 677.75lb = 678lb Ans
2
−
6
∫ x PdA = ∫ x(10)[ ( x + 1) + 8].3dx
0
= 30(6( x − ln( x + 1) + 4 x 2 | 02 = 642.25lb = 678lb
−
−
x=
∫ x. P.dA
P.dA
=
−
642.25
= 0.948 ft , y = 1.50 ft (by symmetry)
677.75
Ans
PAGE
14
14