This preview shows pages 1–3. Sign up to view the full content.
MAE 101: Statics and Strength of Materials
September 27, 2011
Problem 1
The unstretched length of the spring
AB
is 660 mm, and the spring constant
k
= 1000 N/m. What
is the mass of the suspended object?
350 mm
B
400 mm
600 mm
A
Solution:
Use the linear spring forceextension relation to ﬁnd the magnitude of the tension in spring
AB
.
Isolate juncture
A
. The forces are the weight and the tensions in the cables. The angles are
tan
α
=
(
350
600
)
= 0
.
5833
, α
= 30
.
26
◦
(1)
tan
β
=
(
350
400
)
= 0
.
875
, β
= 41
.
2
◦
(2)
The angle between the
x
axis and the spring is
α
. The tension is
T
AB
=

T
AB

(
i
cos
α
+
j
sin
α
)
.
The angle between the
x
axis and
AC
is (180

β
). The tension is
T
AC
=

T
AC

(
i
cos(180

β
) +
j
sin(180

β
))
T
AC
=

T
AC

(

i
cos
β
+
j
sin
β
)
.
The weight is:
W
= 0
i
 
W

j
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document350 mm
B
400 mm
600 mm
A
x
y
C
A
B
`
_
W
The equilibrium conditions:
X
F
=
W
+
T
AB
+
T
AC
= 0
.
Substitute and collect like terms
X
F
x
= (

T
AB

cos
α
 
T
AC

cos
β
)
i
= 0
,
X
F
y
= (

T
AC

sin
α
+

T
AB

sin
β
 
W

)
j
= 0
.
Solve:

T
AC

=
±
cos
α
cos
β
²

T
AB

and

W

=
±
sin 2
α
+sin 2
β
2 cos
β
²

T
AB

.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 ORIENT
 Statics

Click to edit the document details