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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

        Fnet = 0 N :  velocity is constant

 

2.       Second Law of Newton

Fnet= m.a

        Fnet 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 m3

ρ    density in kg m-3

 

Example  

Find the volume of 25 g gold.

ρ = 19.3 x 103 kg m-3     m = 25 g = 0.025 kg

ρ = m/V     19.3 x 103 = 0.025/V      (V) (19.3 x 103) = 0.025      V = 0.025/(19.3 x 103) = 1.30 x 10-6 m3=

1.30 cm3

 

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

Fc = mv2/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.

 

Fc = mv2/r = (0.100)( 7.39) 2 /0.20  = 27.3 N

or Fc = m ac = (0.100)( 273.06) = 27.3 N

For ac : 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.

Fn  normal force

W   weight

f    frictional force

 

Fn  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 Fc  is the  gravitational force Fg

 

Important formulas      Fc= (mv2)/r

                                         Fg = G( M m) /r2

                                                                v = 2 π r /T

 

m  mass of the satellite in kg

M  mass of the earth   ( Mearth = 5.976 x 1024 kg)

G  gravitational constant in N m2kg-2  ( G= 6.6726 x 10-11 Nm2 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 ?

 Fc= Fg

mv2/r = G (m M)/r2

v2 = G M/r         

r = G M/v2    (1)

       v = 2 π r /T (2)

 

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

r3 = (GM T2)/4 π2

r3 = ( 6.6726 x 10-11)( 5.976 x 1024) (24 x 3600)2)/4 π2 = 7.5400 x 1022

r = 4.217 x 107 m = 42.17 x 103  km

 

Altitude satellite = r – Rearth = 42.17 x 103 – 6.378 x 103 = 35.8 x 103 km

 

Find g at this altitude

Fg = m g = (G M m) / r2             

g = Fg/m = (GM)/r2 =( 6.6726 x 10-11)( 5.976 x 1024) / (4.217 x 107)2  = 0.22 m/s2

 

 

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 F1 with respect to P = F1 l 1=  -  (3.0)( 0.02) =

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

Torque of F2 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 (FA )

 

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 - FA  AS = 0

(30.0)( 0.25) -  (FA) (0.75) = 0

FA = 10 N

 

b.Find the force on the bar in S.

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

FS = W + FA

 

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 30o = Fy/F          Fy = F sin 30o = 0.5  F

 

cos 30o = Fx/F          Fx = F cos 30o = 0.866 F

 

Sum of the  torques is zero

Σ τ = 0    with respect to S

FA  AS - Fy  BS = 0

 

(10)( 0.50) – (Fy )( 40)= 0

 

Fy = 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:  FSX  = 30.0 – 25.0 = 5.0 N  (upward)

In horizontal direction   : FSX = F 0.866    FSX = (25 )( 0.866) = 21.65 N (to the left)

Both components of Fs are known:

 

 

 

 FS = 22,2 N   

 ( Pythagorean theorem   FS2= FSX2 + FSY2)

 

Calculation of the direction :

tan α = 5.0/21.65 =  0.2309

 

α = 13 o