Dynamics: Inclines & Machines

A. Forces on INCLINED Inclined Planes:

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Free Body Diagrams    (F.B.D.)
A free body diagram is a simple sketch showing all the forces acting on an object when it is on its own (i.e. removed from its surrounding).  Here are some guidelines for drawing Free Body Diagrams.

Also see FORCES

A force is usually described as a vector quantity with a definite magnitude and a definite direction which can be either a push or a pull on an object.


VECTORS

Also see Components of Vectors

Any vector is space can be defined by its coordinates in terms of its x, y, and z position.

In a two dimensional frame we need only consider its x and y co-ordinates.  Therefore any vector can be thought of as a resultant when its horizontal component in the x direction is added to its vertical component in the y direction.

The  component of a vector V in standard position is Vx  and the y component is Vy.

Where:

Vx = V cosQ       and      Vy = VsinQ

Then we can define V as

v = vx + vy


Gravity & Weight

The force of gravity as we know id expressed as Fg = mg.  Where g is the acceleration due to gravity close to the surface of the earth is (9.8 m/s2 ).  A "more common" name for force of gravity is Weight.  Weight should not be substituted as a synonym for mass.

The acceleration of an object down an inclined plane (a) is equal to the value of the acceleration due to gravity (g) diluted by a factor corresponding to the sine of the angle of the incline with respect to the horizontal (sinq).

a = g x sinq


Friction

Friction is generally defined as the "force that opposes" motion.

When the applied force on an object is equal or less than the force of friction, the object will not move.  If the applied force is slightly bigger than the force of friction, the object will move in the direction of the applied force.  There are two types friction forces  acting on the object.

Static Friction is larger than Kinetic (moving) friction.

If an object is moving at constant speed , it will slow down and eventually come to a stop because the force of friction is constantly acting against it.

In our laboratory experiments we  found that friction depends on several factors:

  1. The weight of the object i.e. the force of gravity ( Fg). Recall that Fg = m x g

  2. Hence indirectly friction depends on mass 

  3. Friction also depends on the type of surface

  4. The presence of a lubricant

The general equation for calculating friction is   Ff = mmg
Where m is the coefficient of friction, m is the mass and g is 9.8 m/s2


Sample Problem: The inclined Plane

A skier goes down a smooth 30o hill (frictionless) for a distance of 10 m to the bottom of the hill where he then continues on a frozen, frictionless pond.  After that, he goes up a hill inclined at 25o  to the horizontal.  How far up this second hill does he go before stopping if the coefficient of friction on this hill is 0.10?

 

Analysis and Solution:

Part I - Accelerated Motion

a = (g)(sin30o)
  
 = 9.8 X 0.5 m/s2

V1 = 0     ( V2 ) 2 =  ( V2 ) 1 + 2(a)(d)      d = 10 m

 ( V2 )     = 9.9 m/s

Part II - No acceleration

Therefore  Fnet = ma = 0  and Vis the same as Vfrom Part 1.

Part III - Friction up an inclined plane

From the Free Body Diagram

Fnet = -Fgx - F

ma =  -(m)(g)(sin 25o) - (m)(g)(cos 25o)

m drops off on both sides of the equation

we solve for the acceleration, a

a = - 4.99 m/s2

we use the acceleration to find the distance, d


Where V2  is zero (skier stops), Solving for d, where V1 = 9.9 m/s from part 1, and  a = -4.99 m/s

We obtain a value for d:

d = 9.8 m

Therefore the distance traveled by the skier before he stops up the second hill is 9.8 m

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