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ME 270 – Fall 2020 Exam 2 SOLUTION
PROBLEM 1 (20 points) Part 1A – 7 points A block of weight W and
center of mass at G is acted upon by two applied forces, P and 3P,
as shown, where P = W/4. The block rests on a rough, horizontal
surface. (a) On the artwork provided below right, draw a free
body
diagram of the block. (1 pt) (b) Let d be the distance of the
normal contact force on
the bottom of the block as measured from lower left-hand corner
A of the block. Determine the value of d for equilibrium. Express
your answer in terms of b.
Fy = −W +N =0 ⇒ N =W = 4P∑MA∑ = − 3P( ) 2b( )−Wb+P 4b( )+Nd =0
⇒
d =2P +W( )N
b=2P +4P( )4P b=
32b
d = 3b/2 (2 pts)
(c) If the block is in static equilibrium, determine the force
of friction f on the block by
the horizontal surface on which it rests. Express your answer in
terms of P. Fx =3P −P − f =0 ⇒ f =2P∑
f = 2P (2 pts)
(d) For a value of coefficient of static friction of µS = 0.6
and for P = W/4, circle the
response below that most accurately describes the motion of the
block: (2 pts) i. The block is at rest. ii. The block is in a state
of impending sliding. iii. The block is in a state of impending
tipping. iv. The block has tipped. v. The block has slipped in the
horizontal surface.
fmax = µSN = 0.6( )4P =2.4P > f ⇒ does NOT slip Since d
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ME 270 – Fall 2020 Exam 2 SOLUTION
P
smoothbearing
µS =0.3
µS =0.2
µS =0.6
W
A
B
C
g
stationary drum
stationary drum
P
A
C TACTAC
W
IMIM
P
µS
Wnegligible weight
A
B
g
B B
NBfB
NAB
fAB
P
IM
B
NBfB
NAB
fAB
P
IM
PROBLEM 1 (continued) Part 1B – 4 points A single cable is
wrapped around two fixed drums, A and C, and a pulley B. A block
having a weight of W is to be held in place with a force P applied
to the other end of the cable. All free sections of the cable are
either vertically or horizontally oriented. Determine the smallest
value of P for which the system can remain in equilibrium. Express
your answer in terms of W.
TACP
= e 0.6( ) π/2( ) = e 0.3π( )
WTAC
= e 0.2( ) π/2( ) = e 0.1π( )
Therefore:
WTAC
⎛
⎝⎜
⎞
⎠⎟TACP
⎛
⎝⎜
⎞
⎠⎟ =
WP
= e 0.1π( )e 0.3π( ) = e 0.4π( ) ⇒ Pmin = e− 0.4π( )W
Pmin = e− 0.4π( )W (4 pts)
Part 1C – 4 points Block A, of weight W, is supported by wedge B
(having negligible weight). A force P acts to the left on the
wedge.
(a) Draw the FBD below of the wedge corresponding to the
smallest value of P for which the wedge is held in place. Be sure
to show the friction forces on the wedge in the correct directions.
Label your forces. (2 pts)
(b) Draw the FBD below of the wedge corresponding to the largest
value of P for which the wedge is held in place. Be sure to show
the friction forces on the wedge in the correct directions. Label
your forces. (2 pts)
P
smoothbearing
µS =0.3
µS =0.2
µS =0.6
W
A
B
C
g
stationary drum
stationary drum
P
µS
Wnegligible weight
A
B
g
B
P
µS
Wnegligible weight
A
B
g
B B
NBfB
NAB
fAB
P
IM
B
NBfB
NAB
fAB
P
IM
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ME 270 – Fall 2020 Exam 2 SOLUTION
A
B
gR
O
O A
B
deq ,V
deq ,H
Feq ,V
Feq ,H
A
B
gR
O
O A
B
deq ,V
deq ,H
Feq ,V
Feq ,H
B
O AW =wt .of water
ρgRb
PROBLEM 1 (continued) Part 1D – 4 points A quarter-circle water
gate is pinned to ground at B, and is supported by a roller at A.
The gate, having a dimension into the page of b, holds back water
having a mass density of ρ . The figure below on the right
indicates the horizontal and vertical components of the hydrostatic
loads on the gate,
Feq ,H and Feq ,V , respectively. The distances deq ,H and deq
,Vrepresent the distances from the center of the gate O to the
lines of action of Feq ,H and Feq ,V , respectively. Determine Feq
,H ,Feq ,V , deq ,H and deq ,V in terms of, at most, g, b, R and ρ
. Include your answers in the answer boxes below.
Feq ,V =weight of water above gate = ρg
14πR
2⎛⎝⎜
⎞⎠⎟b= 14πρgR
2b
deq ,V =
4R3π
Feq ,H = equiv.hydrostatic load =
12 ρgRb( )R =
12ρgR
2b
deq ,H =
23R
Feq ,H =
14πρgR
2b (1 pt)
deq ,V =
4R3π (1 pt)
Feq ,H =
12ρgR
2b (1 pt)
deq ,H =
23R (1 pt)
A
B
gR
O
O A
B
deq ,V
deq ,H
Feq ,V
Feq ,H
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ME 270 – Fall 2020 Exam 2 SOLUTION
x
y
A B C
D
EG
H
I
JKM LP 2P
P
Mx
M y
Ax
PROBLEM 2 (20 points) GIVEN: The truss to the right is loaded as
shown. The truss is supported by a pin joint at M and by a roller
support at A. FIND: For this problem, complete the following. Part
2A – 1 point On the artwork provided to below right, complete the
overall free body diagram of the truss. Part 2B – 3 points
Determine the support reactions on the truss at A and M. State your
answers in terms of P, and as vectors in terms of their
xy-components.
MM∑ = −P 6d( )− 2P( ) 9d( )−P 12d( )+ Ax 4d( ) =0 ⇒Ax = 9P
Fx = Ax +Mx −P∑ =0 ⇒ Mx =8P
Fy =My −P −2P =0 ⇒ My =3P∑
!A = 9Pî (1 pt)
!M =8Pî +3Pĵ (2 pts) Part 2C – 4 points Identify all
zero-force members. No work needs to be shown here for full
credit.
List the zero-force members: AM, CI, DH, GH (4 pts)
x
y 2P
A B C
D
EG
H
I
KM L
P
3d
3d
3d
3d3d3d
3d3d
4d4d
4d4d
4d4d4d4d
J
P
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ME 270 – Fall 2020 Exam 2 SOLUTION
C
D
EG
H
I
P
cut
FIJ
FCI =0
FCH
FCD
θ
cosθ = 35sinθ = 45
C
D
EG
H
I
P
cut
FIJ
FCI =0
FCH
FCD
θ
cosθ = 35sinθ = 45
J
2PFKJ
FIJ
FCJ
θ
PROBLEM 2 (continued) Part 2D – 8 points Using the method of
sections, determine the magnitudes of the loads carried by members
CH and IJ. State your answers as magnitudes in terms of P, and
state whether each member is in tension or compression.
MC∑ = −P 8d( )+FIJ 3d( ) =0 ⇒ FIJ = 83P T( )Fx∑ = −P −FCHcosθ =0
⇒ FCH = −
Pcosθ
= −53P C( )
FCH =5P /3 compression (4 pts)
FIJ =8P /3 tension (4 pts) Part 2E – 4 points Using the method
of joints, determine the magnitude of the load carried by member
CJ. State your answer as a magnitude in terms of P, and state
whether the member is in tension or compression.
Fy∑ = −2P −FIJ −FCJsinθ =0 ⇒
FCJ = −2P +FIJsinθ
⎛
⎝⎜⎜
⎞
⎠⎟⎟= −54
143 P
⎛⎝⎜
⎞⎠⎟= −353 P C( )
FCJ =35P /3 compression (4 pts)
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ME 270 – Fall 2020 Exam 2 SOLUTION
Prb_3_rev Page 1
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ME 270 – Fall 2020 Exam 2 SOLUTION
