A large concrete dam is to be constructed. The center section of the dam is 1200 m wide
and 200 m tall. The upstream face of the dam slopes at an angle of 60 degrees from the
horizontal.
Calculate the horizontal force acting on the concrete dam.
What is the elevation (measured from the base of the dam) of the horizontal force line of
action?
Calculate the vertical force acting on the dam. Note that the horizontal force tends to
destabilize the dam by causing it to want to slide downstream and the vertical force makes
the dam more stable. Should the angle (θ) be increased or decreased to increase the vertical
force?
kg
h := 200m
ρ := 1000
3
m
θ
w := 1200m
p c :=
h
2
⋅ρ⋅g
200 m
FH := p c⋅ h ⋅ w
2
FH :=
ρ ⋅ g⋅ h ⋅ w
2
FH = 2.354 × 10
y R :=
y R :=
11
N
−ρ ⋅ g ⋅ cos( θ ) ⋅ Ixc
Note that this θ is not the same as the θ in the drawing!
p c⋅ A
The projected surface is vertical and gcosθ is g.
2
Ixc
h
:=
12
A
−ρ ⋅ g ⋅ Ixc
p c⋅ A
y R :=
−ρ ⋅ g ⋅ h
y R :=
2
h⋅ ρ ⋅ g⋅ 6
−h
6
elevation :=
h
elevation :=
h
2
−
h
6
3
elevation = 66.667 m
line of action for horizontal force
Vertical force is weight of water above the dam
Let b = the horizontal projection of the dam
h
tan( θ ) =
θ := 60deg
b
b :=
h
b = 115.47 m
tan( θ )
Volumewater :=
h
⋅
⋅w
2 tan( θ )
h
Weight := Volumewater⋅ ρ ⋅ g
2
Weight :=
w⋅ h ⋅ ρ ⋅ g
2 tan( θ )
Weight = 1.359 × 10
11
N
increase b to increase weight above dam. Therefore decrease the angle between the
water surface and the dam.
The braking force of bicycle brakes decreases significantly when the bicycle rims are wet. The
braking force of 4 brake pads that measure 0.5 cm x 4 cm when each brake pad is pressed
with a force of 200 N is only 15 N when the bicycle is traveling at 10m/s. Note that the bicycle
rim velocity relative to the brake pads is almost the same as the bicycle velocity relative to the
ground! Estimate the thickness of the water layer between the brake pads and the wheel rim.
You may assume the velocity gradient between the brake pad and the wheel rim is linear and
constant thickness.
N⋅ s
w := 0.5cm
µ := 0.001 ⋅
2
m
l := 4cm
Abrake := 4 ⋅ w⋅ l
Abrake = 8 cm
2
Fbrake := 15N
τ := µ ⋅
∆V := 10
∆V
∆y
Fbrake = µ ⋅
∆V
∆y := µ ⋅ ∆V⋅
∆y
⋅ Abrake
Abrake
Fbrake
∆y = 0.533 µm
∆y := µ ⋅ ∆V⋅
4 ⋅ w⋅ l
Fbrake
m
s
A ring of diameter 5 cm is gently lifted from the surface of a pool of water and the force
required to lift the ring is carefully measured. The maximum force occurs right before the
surface film breaks and the water surface is vertical.
d := 5cm
Fsurfacetension := 0.02N
because there is an inner and an outer ring
L := 2 ⋅ π ⋅ d
Fsurfacetension = σ ⋅ L
σ :=
Fsurfacetension
σ = 0.064
2⋅ π ⋅ d
N
m
where L is the length over which the force is acting