N ID DATE MTE 119 S P HOMEWORK 9 1 S 14

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 1
Mechatronics Engineering SOLUTIONS
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PROBLEM 8-39
Two blocks A and B have a weight of
SOLUTION 8-39: 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.

(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):

∑ Fx = 0 FA −10sinθ = 0 (1)
∑ Fy = 0 N A −10 cosθ = 0 (2)

From FBD (b): (3)
(4)
∑ Fx = 0 FB − 6sinθ = 0
∑ Fy = 0 N B − 6 cosθ = 0

Friction: Assuming block A is on the verge of slipping, then:

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 2
Mechatronics Engineering SOLUTIONS
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FA = μsA N A = 0.15N 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 = μsB 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 = μC = 0.5 and μ D = 0.6 ,

determine the couple moment M
needed to rotate cylinder E.

SOLUTION 8-46:

(a) (b)

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 3
Mechatronics Engineering SOLUTIONS
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FBD (a):

∑ Fx = 0 N D − FC = 0 (1)

∑ Fy = 0 NC + FD − 490.5 = 0 (2)

∑ M O = 0 M − FC (0.3) − FD (0.3) = 0 (3)

FBDD (b):

∑ Fx = 0 N A − FB − N D = 0 (4)

∑ Fy = 0 N B + FA − FD − 490.5 = 0 (5)

∑ M P = 0 FA (0.3) − FB (0.3) − FD (0.3) = 0 (6)

Friction: Assuming cylinder E slips at points C and D and cylinder F does not move, then:

FC = μsC NC = 0.5NC and FD = μsD N D = 0.6N D . Substituting these values into

Equations (1) and (2) and (3) and solving, we have:

NC = 377.31N N D = 188.65N M = 90.6N.m Ans

If cylinder F is on the verge of slipping at point A, then FA = μsA N A = 0.5N A . Substitute

this value into Equations (4), (5) and (6) and solving, we have:

N A = 150.92N N B = 679.15N FB = 37.73N Ans

Since (FB )max = μsB N B = 0.5(679.15) = 339.58N > 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

μS = 0.5 and at B, μ 'S = 0.6 . Determine

the couple moment M needed to rotate
cylinder E.

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 4
Mechatronics Engineering SOLUTIONS
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EXTRA PRACTICE SOLUTION 8-63:

(a) (b)
Equations of equilibrium:

FBD (a):

∑ Fx = 0 N B cos 30D − FB cos 60D − NC = 0 (1)
∑ Fy = 0 N B sin 30D + FB sin 60D + FC − W = 0 (2)

FBD (b): (3)
(4)
∑ Fx = 0 N A − N B sin 30D − FB sin 60D − 8 = 0 (5)
∑ Fy = 0 FA + FB cos 60D − N B cos 30D = 0
∑MO = 0 FA (0.5) − FB (0.5) = 0

Friction: Assume slipping occurs at points C and A, then FC = μsC NC = 0.5NC
and FA = μsA N A = 0.5N A . Substituting these values into Equations (1),(2), (3),(4) and (5)

and solving, we have:

W = 66.64lb Ans

N B = 51.71lb N A = 59.71lb FB = NC = 29.86lb

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 5
Mechatronics Engineering SOLUTIONS
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PROBLEM 8-70
If the beam AD is loaded as
SOLUTION 8-70: 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 μCA = 0.25 and μCB = 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.

(a)

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 6
Mechatronics Engineering SOLUTIONS
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Equations of equilibrium and friction:
(b)

If the wedge is on the verge of moving to the right, then slipping will have to occur at both

contact surfaces. Thus, FA = μsA N A = 0.25N A and FB = μsB N B = 0.35N B .

FBD (a): N A cos10D (7) + 0.25N A sin10D (7) − 6(2) −16(5) = 0
N A = 12.78kN
∑MD =0

FBD (b): N B −12.78sin 80D − 0.25(12.78) sin10D = 0
N B = 13.14kN
∑ Fy = 0

∑ Fx = 0 P + 12.78cos80D − 0.25(12.78) cos10D − 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/ft3 . Determine

the location of the center of gravity. Also,
find the tension in each of the cords used to
support it.

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 7
Mechatronics Engineering SOLUTIONS
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EXTRA PRACTICE SOLUTION 9-13:
y Calculation of the area and the
first moments:

(x,y) Height of the differential element:
16
1
(x, y)
y = x − 8x2 +16
dx
16 Area of the differential element:
⎛ 1⎞

dA = ydx = ⎜ x − 8x2 +16⎟ dx
⎝⎠

x

Centroid of differential element:

x = x and y = 1 ⎛ x 1 ⎞
2 ⎜ + 16 ⎟
⎝ − 8x 2
⎠

The total area is given by,

∫ ∫A= dA = 16 ⎛ x − 1 ⎞ = ⎛ x2 − 16 3 ⎞ 16 = 42.67 ft 2
⎜ + 16 ⎟dx ⎜ 2 3 +16x ⎟ 0
8x2 ⎝ x2
0⎝ ⎠ ⎠
A

Therefore:

xdA 16 ⎡⎛ 1 ⎞⎤ ⎛ x3 − 16 5 + 8x2 16 ft3
∫ ∫A = 0 x ⎢⎜ x +16⎟ dx⎥ = ⎜ 3 5 = 136.53
⎣⎢⎝ − 8x 2 ⎝ x2 ⎞
⎠ ⎥⎦ ⎟
⎠0

ydA 16 1 ⎛ 1 ⎞ ⎡⎛ 1 ⎞⎤
∫ ∫A = 0 2 ⎜ x − + 16 ⎟ ⎢⎜ x +16⎟ dx⎥
⎝ 8x2 ⎣⎢⎝ − 8x 2
⎠ ⎠ ⎥⎦

∫A ydA = 1 ⎛ 1 x3 − 32 5 + 48x2 − 512 3 + 16 = 136.53 ft3
2 ⎜ 3 5 3
⎝ x2 x2 ⎞
256x ⎟

⎠0

The coordinates of the centroid are therefore given by:

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 8
Mechatronics Engineering SOLUTIONS
14

∫ xdx 136.53

x = A = = 3.2 ft ANS

∫ dA 42.67
A

∫ ydx 136.53

y = A = = 3.2 ft ANS

∫ dA 42.67
A

Tension in the cables:
The weight of the plate is: W= 42.67(0.25)(180)=1920 lb

TB
TC

W
TA

Equations of equilibrium: TA = 384 lb ANS
TC = 384 lb ANS
∑ M x = 0; 1920(3.20) − TA (16) = 0 ANS
∑ M y = 0; TC (16) −1920(3.2) = 0

TB = 1152 lb

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.

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 9
Mechatronics Engineering SOLUTIONS
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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 2dz = 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)
m0

= πk( r 2 z 2 − z2 ) |0r = πkr 4 Ans
2 4 4

−r

∫ z dm = ∫ z{kz[(π (r 2 − z 2 )dz]}

m0

= πk( r 2 z 2 − z2 )) |0r = 2πkr 5
3 5 15

Centroid: Applying Equation 9-4, we have:

−

∫ z dm

m = 2πkr 5 /15 = 8 r

∫ dm πkr 4 / 4 15

m

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 10
Mechatronics Engineering SOLUTIONS
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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:

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 11
Mechatronics Engineering SOLUTIONS
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−

∑ x.W = 1.5(4) 45 + 3(4) 72 = 142.073lb. ft
∑ W = 4(7) + 4 45 + 4 72 = 88.774lb

−
∑ x.W
− ∑W = 142.073 = 1.06 ft Ans
88.774
x=

−

∑ y.W = 3.5(4)(7) + 7(4) 45 + 10(4) 72 = 625.241lb. ft

−
∑ y.W
− ∑W = 625.241 = 7.04 ft Ans
88.774
y=

∑ Fx = 0 Ax = 0
∑ Fy = 0 Ay = 88.774 + 60 = 149lb
∑MA =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.

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 12
Mechatronics Engineering SOLUTIONS
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SOLUTION 9-67:

Centroid: The area of each segment and its respective centroid are tabulated below:

Segment A(in2 ) − −

1 14(0.4) y(in) y A ( in )
2 3.4(1.3)
3 10.3(0.76) 16.20 90.72
4 14.48(0.56) 14.7 64.97
5 10.3(0.76) 15.62 122.27
8.00 64.87
∑ 33.78 0.38 2.97
345.81

Thus:

−
∑ y.A
− = ∑A = 345.81 = 10.24in = 10.2in Ans
33.78
y

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 13
Mechatronics Engineering SOLUTIONS
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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 0h = 6(56)(2) = 672lb / ft
WC = bγ 0h = 6(56)(10) = 3360lb / ft
Fh1 = 8(672) = 5376lb
Fh2 = 1/ 2(3360 − 672)(8) = 10752lb
FV1 = (3)(2)(6)(56) = 2016lb
FV 2 = 1/ 2(3)(8)(6)(56) = 4032lb

∑ Fx = 0 FRx = 5376 + 10752 = 16128lb
∑ Fy = 0 FRy = 2016 + 4032 = 6048lb

FR = (16128)2 + (6048)2 = 17225lb = 17.2kip Ans

θ = tan −1 ( 6048 ) = 20.56D
16128

NAME & ID DATE MTE 119 – STATICS PAGE
HOMEWORK 9 14
Mechatronics Engineering SOLUTIONS
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∑ M RB = 0 17225d = 10752(2 / 3)(8) + 5376(4) + 2016(1.5) + 4032(2)
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 6

PdA = 10[
∫ ∫FR = 0 (x + 1) + 8].3dx

FR = 30(6 ln(x + 1) + 8x |02 = 677.75lb = 678lb Ans

∫ ∫− 2 6 + 8].3dx
x PdA = x(10)[
0 (x + 1)

= 30(6( x − ln(x + 1) + 4x2 | 2 = 642.25lb = 678lb
0

−

∫− x. P.dA = 642.25 = 0.948 ft − Ans
P.dA 677.75
x= , y = 1.50 ft (by symmetry)