7.6 Application to Physics and Engineering

7.6 Applications to Physics and Engineering
Objectives:
Three applications:
1. Work
2. Hydrostatic Pressure and Force
3. Moments and Centers of Mass
1. Work
Recall: Work = _________________
(this assumes force is constant)
What happens if force is a variable?
Suppose an object moves along the x-axis in the positive
direction from x = a to x = b and at each point x between a
and b a force f(x) acts on the object where f(x) is
continuous.
Recall Hooke’s Law: The force required to maintain a
spring stretched x units beyond its natural length is
proportional to x.
Ex: A spring has a natural length of 20cm. If a 25-N force
is required to keep it stretched to a length of 30cm, how
much work is required to stretch it from 20cm to 25cm?
(Note: Newton: N = kg m / s 2
1cm = 0.01 m
Joule: J = N-m (Newton-meter) )
Ex: A cable that weight 2 lb/ft is used to lift 800 lbs of
coal up a mine shaft 500 ft deep. Find the work done.
2. Hydrostatic Pressure and Force
Drop a thin horizontal plate with area A square meters in a
fluid of density ρ kg/m 3 at a depth d meters.
• The fluid directly above the plate has volume V = Ad
• Mass: m = ρV = ρAd
• Force: F = mg = ρAdg , where g is acceleration due
to gravity.
• Define pressure: P = F / A = ρdg (force area per
unit), measured in pascals = N/m2
Recall: At any point in a liquid the pressure is the same
in all directions.
• P = ρdg = δd
• We can use this to help us determine the
hydrostatic force against a vertical plate (or wall or
dam) in a fluid. (* not easy: pressure is not
constant, but increases as depth increases)
Ex: A large tank is designed with ends in the shape of the
1 2
region between the curves y = x and y = 12 , measured
2
in feet. Find the hydrostatic force on the end of one end of
the tank if it is filled to a depth of 8 ft with gasoline.
(assume the gasoline’s density is 42 lb/ft3)
3. Moments and Centers of Mass
Defn: The center of mass is the point on which a thin
plate of any given shape balances horizontally.
Consider two masses m1 and m2 attached to a rod of
negligible mass on opposite sides of a fulcrum, at distance
d1 and d 2 . The rod will balance if m1d1 = m2 d 2 (Law of the
Lever).
Now suppose that the rod lies along the x-axis with m1 at
x1 and m2 at x2 , and the center of mass at x . x2
d1 = x − x1 and d 2 = x2 − x
m1d1 = m2 d 2
m1 ( x − x1 ) = m2 ( x2 − x )
Solve for x
Defn: The numbers m1 x1 and m2 x2 are called the moments
of the masses m1 and m2 (with respect to the origin).
In general, if we have a system of n particles with masses
m1 , ..., mn located at the points x1 , ..., xn on the x-axis,
the center of mass of the system is located at:
n
Defn: the sum of the individual moments M = ∑ mi xi is
i =1
call the moment of the system about the origin.
Now move to two-dimension: consider a system of n
particles with masses m1 , ..., mn located at the points
( x1 , y1 ), ..., ( xn , yn ) in the xy-plane.
Defn: the moment of the system about the y-axis is
n
M y = ∑ mi xi , and the moment of the system about the xi =1
n
axis is M x = ∑ mi yi . (to be continued)
i =1

Download Report