Joe fkrpech
ME-211 Statics
Kwok-Choon F22
Homework Assignment #2
Due 10/5/2022
Moments 2D, Moments 3D, Force Equilibrium, Moment Couples
Q1
Q2
Q3
Q4
Q5
Q6
/15
/15
/15
/15
/10
/15
Total
/85
Provide each answer in the following format:
Given: State Given Variables
Find: State Variables to Solve for
Type: State type of problem
Solution: Provide Free Body Diagram and solution method. Box final answer.
Reflection: Discuss whether solution makes sense
Page 1 of 5
ME-211 Statics
Sketch
Kwok-Choon F22
:
_••↑T*
Question 1: (15 points)Calculate the moment that
the 90-N force creates about point O for the
condition θ = 15°. Solve this through two methods.
-
y
-
-
-
1ST
'
-
'
<
=/ 05
)⊖
☆
-90s:nGs)
Mo
×
901054s)
1
relative
'
!
>
forces
moment
and vector
cross
us
0.8%-900545×0.6)J
HAS
Max
:-,
point 0 with pintail
moment but also
and
cross products
vector
products
-90s :-( is)(
0.8
0.6
.
'
↑
i
,
Type :
of
B)
"
(A)
Break the force into components and add the
moment caused by each component. (10 points) Given :
(B)
Use vector cross product to solve this
0=15
problem. (5 points) Find : the moment
-90N
F-
µ,
gone :
Fax
N
Fay
--
✗
=
5-
0.6m
0.8m
-1M¢.im#rs)Xdd=l0-F-T
d=F
90s :-(is )
0=1
90hrs45)
=
.in/1s)(o.6)-Co.8)-90cosl-332SN#
Mjfqos
-
Given
:X,y values
of
A. B. C.
Dato)
13=1200165
directional
Type : partial
Forces
and partial
cos
Find
amount
of tension
each
on
chord
Question 2: (15 points) Three wires are connected at point D, which is located 20 inches below a Tshaped pipe support ABC. Determine the tension in each wire when a 200-lb mass is suspended from
point D as shown.
Fg-IOO.386.088.ir/s?F---FgW
a
I
I
B
F= 77,217.6
¥
i
-
-
-
-
-
-
.
-
(
.
=
/
lbs inches
20
!.it?....#.&....D..-.---.--
'
20
2
•
.
12025
"
e
,
"
B
4
i
•
Fg
-
-
-
-
¥
I
A•
!
•
µ
.
_
Page 2 of 5
BYE
are
"
1
sketch :
Bq
•
Given :
D- Mg
c
f-
.
Ftnet
its
_•!?
.
I
.
.
Find
25
•
: Tension
Type :
A
solution
Reflection
:
tension
sense
=
less
F☐=(200319-8)
F- aunt
ABC
:-
3D with
☒
partial Magnitude
201^+205 -15k
1=20+2%+1-157
lDB%%¥◦!%É=E
,
makes
and
Ñ
Eauaiisruim
N
y
'
lbs
✗
-
20
20
__
,B,c,D
14=200
=
,
15 :
A
for
✗ is ,2
means
work
and
distribution
Eaval
* µ,
ay
They
have
earn tension
§
115131=1154
because
1
i
trig
\
Fy
1
1
gpg.ga.a.g.mg
¥-1
-
I
01^+205 -125k
20¥28
☒F320llb@
=
ME-211 Statics
Kwok-Choon F22
Question 3: (15 points)
A crate of 100 kg is held in position as
shown.
(A)
The moment produced by the weight
W of the crate about E. Solve this by
breaking the force into components and
add the moment caused by each
component. (10 points)
(B)
The smallest force vector exerted at
D that creates an equal but opposite
moment about pivot point E. (5 points)
Question 4: (15 points) Determine the moment about the origin O of the force
̂ (𝑁)
⃗𝑭 = 4𝒊̂ − 3𝒋̂ + 7𝒌
That acts at a point A. Assume the position vector of A is at
(a)
(b)
(c)
̂ (𝑚)
𝑨 = 10𝒊̂ − 10𝒋̂ + 5𝒌
̂ (𝑚)
𝑨 = 2𝒊̂ + 4𝒋̂ − 10𝒌
̂ (𝑚)
𝑨 = −10𝒊̂ + 10𝒋̂ − 5𝒌
(5 points)
(5 points)
(5 points)
Use vector cross product.
Page 3 of 5
Given :X ,y
:
Type
widths
M=
100kg
moment
and
Find : moment
Sheth
1. 2
point
,
C D
,
B A
,
,
forces
parfait
◦
+
☐
moment
it 's
from
force
Eavilibruim
the
and
E
at
crate
of
E
Mlwide)
Reflection : My
makes
n
F"
¥
e-
DX
F
2
☆
10.85)
-
F-
pivot point
farther
from
it
or
require
:S:
-
more
_
force
t.eauaioi-i.tn me
Force Reached
9
the person
by
3- E
D
-
a
would
.ass.s④
^
p
Fw
per
awas
to be held
,
ME -_Ñ•d
=
0.6
i
a)
:( d)
as
sense
.
D
0.15
bing
-12%1=1
*
•
>
answer
(1001/9.8)
Fw -980N
-
µ
)
ME 98010.25
__
=2↳N[m
☆
Sketch :
given
rt
Type
Fri
•
•
10
""
i
II
( g)
4- 37
=
o.io
←
.
•
Mc
b)
*
←
14-3
Lay
1-
.
>
,,
inputs
¢311s) (71-101)/7 -1-[4157-70] I [41-107-130]
-
1%3=130-2811^+1-40 -14)J+( 16-6 )ñ
] Mpg=2i-54J-l0Ñ@
N
~
Reflection
: nah "
Serse
easy matrix
c)
(4-37)
-1010 -5
+
i=55↑-505-1oñ@
9
In
☆
and vector
products
cross
'
a)
-
'
vector
:
solve :
•%
↑=4 J= -3^4=7
:
Me
=
(15-7017+(-20+70))=1/40-30)Ñ
Ñc=-55i+5oJ+10k
ME-211 Statics
Kwok-Choon F22
Question 5: (10 Points) Equivalent Force couple systems.
Decompose the following diagrams into a simplified Force Couple System about Pivot Point A.
° ""
^
"
" ""
+ " ""
"
"
forces
between
and distance
/
Type moment problem
Forces
§
+
,
of
sum
:
Find :
Forces
of
son
point
around
rotating
A
solver :
1PM,+=
(125.07*18011.2)+(-60*2) -11-40×2.0
+
[
EEe+= 125+80-60-40 -_l0④
E-
clockwise
↓
-
2)
MA
(60-0)+(-30*0.05)
-1/50
•
0.557+(140*0.85)
{ F.*
Fret
=
T20
N
•
•
•
>
120N
•
•
-
-
-
-
a
néiters
'
1.81
-
us
-
!
*
T20
N
Reflection
just
pj
isn't
at
applied
:S
is
moment
the
as
direction
-
and
Ms answers
I think
distance
negative
the
:
sense
•
N
105N
*
d
make
0.5
a
'
is
&
n
e.
d=0.5Met①
^
A
sketch
124--1105)•d
Es
MA=Ñ•d
60=120
120N
120
,jѕd
D=-1.181 meters
60-30+50+40
=
M
For Mof system
in
plug
Es
Mµ=60Nm↑p
=
105N
now
counter
-
'
Nm1s2
-124
ME
negative
which
different
Gf
2) Just makes
Sense
Page 4 of 5
ME-211 Statics
Kwok-Choon F22
Question 6 (15 points)
Giver
0--601=-1--1500
:
BC -5m
AB=8m
Type
moment
with
-
problem
Eaonlibruim
:
+
combined Mag
,
Find
to
:
Individual
CÑ
%
and
eauauilent
be
N
combined
Magnitudes
to
1500N
1500N
A hot air balloon is tethered to the ground by a cable attached it is anchor point, B. If the tension in the
cable is 1500 N. Replace the force exerted by the cable at B with an equivalent system by two parallel
forces applied at A and C. "" "
B
A
$-1
solve
8
5m
Reflection
%☆,
*
,
:
g.
:
n
:...→☐
13m
8m
xp
EM -0
of
EFg=
FT -11%
,
=
0
M
1500N
µ
#
Fagin /607
=
Fos:nk0)
=
,
•
Fc
have
×
answer
because
So
,
"
My
sense
makes
,
>
:
;F*
£
a.e.a.se
moments
need
;ffoerf
, :S
+
°
forces
given
f-8)
•
S
r
MB=0
>
¥605
"
→
r
0=-817++517
-
8FA=5Fc
8-
g-
FA=§Fc
N
§ Fc
+
Fc
=/ SOON
E=923.076
FAI
1500
-
923.0769
F*=S76.923
Page 5 of 5