COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 114.
(a)
Let ( R, M D ) be the equivalent force-couple system at D.
First note...
At
x = b; y = h
For
y = k x2
We have h = k b 2
h
or k = 2
b
h 2
∴ y = 2 x
b
h
y = 2 x2
For any contact point c alone the surface
b
dy
h
=2 2x
dx
b
b2
tan −1
2hx
R = F
and
ΣΜ D : M D = − ( x ) F sin θ + ( h − y ) F cosθ
b2
= −x F
b 4 + 4h 2 x 2
or
MD
2h x
+ h − h x 2 F
2
4
b
b + 4h 2 x 2
h 2
2
− xb + h − 2 x ( 2hx )
b
= F
4
2 2
b + 4h x
2h 2 x 3
2
2
− xb + 2h x −
b2
= F
b 4 + 4h 2 x 2
or M D
.
2h 2 x3
2
2
2h − b x −
b2
= F
b 4 + 4h 2 x 2
(
)
COSMOS: Complete Online Solutions Manual Organization System
b = 1 ft, h = 2 ft
(b) With
7 x − 8 x3
MD =
F
2
1 + 16 x
For M D to be a maximum
Then
dM D
dx
7 − 24 x 2
= 0 = F
(
)
1
1 + 16 x 2 − 7 x − 8x3 ( 32 x ) 1 + 16 x 2
2
2
1 + 16 x
(
(
)
)
(
)
−
1
2
For the non-trivial solution:
(
)(
)
(
0 = 7 − 24 x 2 1 + 16 x 2 − 16 x 7 x − 8 x
)
0 = 256 x 4 + 24 x 2 − 7
Solving by the quadratic formula gives for the positive root.
x = 0.354 ft
.
COSMOS Chapter 3 Solution 114
of 2
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