Download Splitting Strategy: 2 Dimensional Problems The first concern in

Splitting Strategy: 2 Dimensional Problems
The first concern in solving 2-dimensional problems is dividing them into two 1dimensional problems we can solve with our tools. Here is a general procedure for
approaching these problems.
A. Choosing directions for coordinates.
Non-circular motion:
Determine what direction the object will accelerate (or move);
choose one axis in the direction of the acceleration,
the other coordinate axis perpendicular to it.
Circular motion:
Choose one coordinate to be radial from the center,
Choose the other axis to be tangent to the circle (angular)
B. Breaking vectors into components along the axes.
Resolve any vectors not along the axes into components on the axes using
sine and cosine.
Vectors which may be split include velocity, force, displacement,
momentum and others we will later encounter.
Example:
C. Use the values along each axis to define a one dimensional problem and apply
the appropriate physical ideas to construct the solution.
Example 1: Motion down a ramp
First: Draw a diagram with known
quantities labeled
A.
Choose coordinates
(gray on diagram)
B.
Resolve vectors not on axes
(mg in this case)
C.
Redraw diagram using components
Divide problem into 1-dimensional problems.
.Along ramp
perpendicular
ax = ?
Force: mg sin() – f = max
Friction: f =  Fn
mg sin() –  Fn = max
Use Fn = mg cos() from other direction
mg sin() –  mg cos() = max
ay = 0 no perpendicular motion
Force: Fn – mg cos(() =may = 0
Fn – mg cos() = 0
Fn = mg cos()
NOW
we look at what the problem actually wants us to find. Here are some
examples:
a) If the 2kg block is sliding down the ramp at constant speed, calculate the coefficient
of kinetic friction if the angle θ = 30º.
b) The ramp is tilted until the block just begins to slide at angle θ . Calculate the
coefficient of static friction.
c) The force of friction acting on the 3kg block is 6N. Calculate the acceleration of the
block down the ramp tilted 30º above horizontal.
Problems (a) and (b) can be solved using mg sin() –  mg cos() = max,
and (c) with mg sin() – f = max