FE Physics

Formulas & Explanation

Formulas & Explanation

Dutch version

**
Force and Torque**

**
The First Two Laws of Newton**

1.
First Law of Newton

A object continues in a
state of rest or in a state of motion, when the net force on that object is zero

F_{net }= 0 N :
velocity is constant

2.
Second Law of Newton

F_{net}= m.a

F_{net }in N

m in kg

a in m s^{-2}

**Mass and Density **

Density is mass per
volume-unit

ρ
= m/V

m
mass in kg

V
volume in m^{3}

ρ
density in kg m^{-3}

*
Example *

Find the volume of 25 g
gold.

ρ
= 19.3 x 10^{3 }kg m^{-3 } m
= 25 g = 0.025 kg

ρ
= m/V 19.3 x 10^{3
}= 0.025/V
(V) (19.3 x 10^{3}) = 0.025
V = 0.025/(19.3 x 10^{3}) = 1.30 x 10^{-6 }m^{3}=

1.30 cm^{3}

**
Weight and Mass**

Mass is independant of the
place

The weight depends on the
local gravitational acceleration g.

W_{ }= m g

On earth’s surface
the gravitational acceleration
is about 9.81 m s^{-2}

The weight of an object with
a mass of 600 g:

W= m g = (0.6)( 9.81) = 5.89
N

**
Uniform Circular Motion-
Centripetal Force**

For a uniform
circular a centripetal force is required

F_{c
}= mv^{2}/r

*
Example *

A disc
rotates in a horizontal plane
with 360 revolutions
per minute

At a distance of 20 cm from
the center is a block with a mass of 100 g.

Find the centripetal force
on the block.

F_{c }=
mv^{2}/r = (0.100)( 7.39)^{
2}
/0.20
= 27.3 N

or F_{c}
= m a_{c}
= (0.100)( 273.06) = 27.3 N

For a_{c
}: see the previous
chapter (Two dimensional motions)

*
Remark*

This centripetal
force is not a new force but it is the required net force and is directed to the
center of the path.

On the block
work 3 forces.

F_{n}
normal force

W
weight

f _{
}frictional
force_{}

F_{n
}
and
W cancel each other

The frictional force f_{
}is the required
centripetal force

**
The Orbit of a Satellite
around the Earth**

The required centripetal
force F_{c }_{
}is
the _{
}gravitational force F_{g
}

Important formulas
F_{c}= (mv^{2})/r

F_{g}
= G( M m) /r^{2}

^{
}v
= 2 π
r /T

m
mass of the satellite in kg

M
mass of the earth ( M_{earth
}= 5.976 x 10^{24}
kg)

G
gravitational constant in N m^{2}kg^{-2
}( G= 6.6726 x 10^{-11} Nm^{2} kg^{-2})

T
period of the satellite in s

r distance between the center of
mass of the earth and the satellite in m

The satellite is in a uniform circular motion.

Geostationary satellites always appear stationary at the same point above the
equator

The period is equal to the rotational period of the earth (= 24 h)

*Example *

What is the height of a geostationary satellite ?

F_{c= }F_{g}

mv^{2}/r = G (m M)/r^{2}

v^{2} = G M/r

r = G M/v^{2
}(1)

^{
}v = 2 π r /T (2)

We substitute v from equation (2) in equation (1) :

r^{3 }= (GM T^{2})/4
π^{2 }

r^{3}
= ( 6.6726 x 10^{-11})(
5.976 x 10^{24}) (24 x 3600)^{2})/4
π^{2
}= 7.5400 x 10^{22}

r = 4.217 x 10^{7 }m = 42.17 x 10^{3}
km

Altitude satellite = r – R_{earth }= 42.17 x 10^{3} – 6.378 x 10^{3}
= 35.8 x 10^{3} km

Find g at this altitude

F_{g}
= m g = (G M m) / r^{2}

g = F_{g/}m = (GM)/r^{2}
=( 6.6726 x 10^{-11})( 5.976 x 10^{24})
/ (4.217 x 10^{7})^{2} =
0.22 m/s^{2}

**Torque of a force with respect to a
point P**

τ = F
r

τ
torque in Nm

F
force in N

l
lever arm (perpendicular distance
from P tot the line of action of F)

If F tends to cause
counterclockwise rotation around P
then we choose
τ > 0

If F tends to cause
clockwise rotation around P then
τ < 0

*Example 1
*

*
*

Torque of F_{1} with
respect to P = F_{1 }
l
_{1}= -
(3.0)( 0.02) =

- 0.06 Nm
(if
l
_{1 }in cm then 6.0 Ncm )

Torque of F_{2 }with
respect to P = 0.06 Nm

*Example 2*

An uniform bar of 2.00 m
with a weight of 30.0 N is supported in
S. The bar can rotate about S in
the vertical plane. The center of mass
O is in the middle.

OS = 25 cm
(see drawing)

The bar is in equilibrium.

a.Find the force on the end
A of the bar (F_{A })

There are two conditions of equilibrium:

1.Σ
τ = 0
with respect to every point

2.ΣF = 0 N

The sum of the torques must be zero

Σ
τ = 0
with respect to S:

W
OS - F_{A} AS = 0

(30.0)( 0.25) -
(F_{A}) (0.75) = 0

F_{A}
= 10 N

b.Find the force on the bar
in S.

The sum of the forces on the
bar must be also zero.

F_{S }= W + F_{A}

*Example 3
*

Given a uniform bar. Lenght = 1.00 m.
The weight of the bar is 20.0 N

A force of 10.0 N acts
perpendicular downward on the bar in A.

The bar is supported in S.

The bar is in equilibrium.

In B, at 40 cm from the
center of mass, a force F acts upwards on the bar at an angle of 30 ^{o}.

a.Find F.

Resolve F into two directions, along the
bar and perpendicular to the bar

Sin 30^{o} = F_{y}/F
F_{y} = F sin 30^{o} = 0.5
F

cos 30^{o} = F_{x}/F
F_{x} = F cos 30^{o }= 0.866 F

Sum of the
torques is zero

Σ
τ = 0
with respect to S

F_{A}
AS - F_{y
}BS = 0

(10)( 0.50) – (F_{y
})( 40)= 0

F_{y }= 12.5 N

Now we can calculate F

F _{y}
= 0.5 F

F = 12.5/0.5 = 25.0 N

b.Find magnitude and
direction of the force that acts on
the bar in S.

A second condition for
equilibrium is

ΣF = 0 N

At first in the vertical
direction: F_{SX
}=
30.0 – 25.0 = 5.0 N (upward)

In horizontal direction
: F_{SX} = F 0.866
F_{SX} = (25 )( 0.866) = 21.65 N (to the left)

Both components of F_{s }are known:

F_{S
}= 22,2 N

( Pythagorean theorem
F_{S}^{2}= F_{SX}^{2} + F_{SY}^{2})

Calculation of the direction
:

tan
α =
5.0/21.65 = 0.2309

α =
13 ^{o}