Transcript Objects on Inclined Planes - Mounds View School Websites

Objects on Inclined Planes
I guess you could call them ramps
if you want.
Inclined Plane
Rotate the axis
y
Rotate the axis so that
the “X” dimension is
parallel to the surface
and the “Y” dimension
is perpendicular to the
surface.
x
Inclined Plane
Rotate the axis
This way the motion
will be in the “X” and
we are back to 1-D
physics!
Draw your free body diagram.
1. Sketch the center of gravity
2. Gravity?
Fg is always directly downward
3. Surface?
Sketch in the surface and
the coordinate system
Fn is always perpendicular
to the surface
4. Friction
Ff is always against the
motion
Fg
5. Other forces?
Get the values.
1.Force due to
Gravity
Fgy = Fg cos q
2. Break up Fg into its
components Fgx and Fgy
3. Normal force
Usually
Fgx = Fg sin q
Fg= m (9.8 N/Kg)
4. Friction
Example
A 10 Kg block is resting on a
35o hill. Find the magnitude
of all of the forces acting on
the block.
Example
A 10 Kg block is resting on a
35o hill. Find the magnitude
of all of the forces acting on
the block.
Ff = -Fgx
Fn = -Fgy
Fgx = 98 N sin 35o
Fg = (10 Kg)(9.8 N/Kg)
Example
A 10 Kg block is resting on a
35o hill. Find the magnitude
of all of the forces acting on
the block.
Ff = -Fgx
Fn = -Fgy
Fg = 98 N
Fgy = - 80.3 N
Fgx = 56.2 N
Fn = 80.3 N
Ff = -56.2N
Fgx = 98 N sin 35o
Fg = (10 Kg)(9.8 N/Kg)
Fgy = - 80.3 N
Example
Fgx = 56.2 N
Fn = 80.3 N
Ff = -Fgx
Fn = -Fgy
Ff = -56.2N
All of the forces are
balanced (SF = 0 ) so
this block will not
accelerate.
Fgx = 98 N sin 35o
Fg = (10 Kg)(9.8 N/Kg)
A 3.0 Kg ball is on a 40o ramp. Find all of the
forces acting on the ball and the acceleration of
the ball. Neglect friction.
Fg = 29.4 N
Fgy = -22.5 N
Fgx = 18.9 N
Fn = - Fgy = 22.5 N
Fgx is unbalanced. So SF = ma
Fgx = ma, 18.9 N = (3.0 Kg) a
a = 18.9 N/ 3.0 Kg
a = 6.3 m/s2
Fg = 3.0 Kg(9.8 N/Kg)