Environmental Engineering Reference
In-Depth Information
vectors
F
max
and
N
that are perpendicular to each other; Eq. (2.29) written with
F
max
and
N
as vectors would incorrectly imply that these vectors were parallel.
Example 2.3.4
A wooden block is at rest on a horizontal wooden plank. One end
of the plank is slowly raised. Determine the angle that the plank makes with the
horizontal when the block begins to slide. The coefficient of static friction between
the block and the plank is 0.3.
Solution 2.3.4
Let the mass of the block be m. The weight of the block has
components mg
sin
θ parallel, and mg
cos
θ perpendicular, to the surface of
the plank. Static friction opposes the force down the slope and has a maximum
value µ
s
mg
cos
θ . When the component of the weight down the slope reaches this
maximum value, the block begins to slide. This occurs at an angle θ
m
where
mg
sin
θ
m
=
µ
s
mg
cos
θ
m
.
Hence
tan
−
1
µ
s
.
θ
m
=
16
.
7
◦
.
Using µ
s
=
0
.
3
this gives θ
m
=
Electrostatic forces between surfaces are of crucial importance to everyday life.
That you are able to stand on the floor is only possible because a condition of static
equilibrium exists with the normal force counteracting your weight. You are able
to write only if the maximum value of static friction between your fingers and the
pen exceeds the force that the writing surface exerts on the pen in a direction along
its length. In some circumstances we try to reduce friction to facilitate motion -
the application of oil to a rusty lock, for example. In other situations friction is
essential to motion: when we walk, it is the friction between the floor and our
feet that causes us to accelerate horizontally. The maximum horizontal force that
we can apply to the floor is equal in magnitude but opposite in direction to the
maximum horizontal force that the floor can apply to our feet and this force is
limited by the maximum static friction. Therefore, running shoes and car tires are
both designed to produce large static friction so that the runner or car engine may
exert large forces on the ground without slipping.
Example 2.3.5
Calculate the maximum acceleration that a four-wheel drive car
may achieve going (a) straight uphill or (b) straight downhill, on a slope that makes
an angle θ with the horizontal.
Solution 2.3.5
As with the block in the previous example, the weight will have
components mg
sin
θ and mg
cos
θ parallel and perpendicular to the slope. Since
there is no component of acceleration perpendicular to the slope, the normal force
must have magnitude mg
cos
θ and the maximum value of static friction will be
µ
s
mg
cos
θ . This is the maximum force that the engine can exert without causing
the wheels to spin. Applying the Second Law to components parallel to the slope,
and taking acceleration uphill as positive gives
=
−
ma
mg(µ
s
cos
θ
sin
θ)
