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Inclined Plane x y Rotate the axis This way the motion will be in the “X” and we are back to 1-D physics!
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Objects on Inclined Planes
I guess you could call them ramps if you want.
Inclined PlaneObject on an inclined plane
x
y
x
y
Rotate the axis
Rotate the axis so that the “X” dimension is parallel to the surface and the “Y” dimension is perpendicular to the surface.
Inclined Plane
x
y
Rotate the axis
This way the motion will be in the “X” and we are back to 1-D physics!
2. Gravity?Fg is always directly downward
Draw your free body diagram.
Fg
3. Surface?Sketch in the surface and the coordinate system
Fn is always perpendicular to the surface
Fn
4. FrictionFf is always against the motion
Ff1. Sketch the center of gravity
5. Other forces?x
y
1.Force due to Gravity
Get the values.
Fg= m (9.8 N/Kg)
3. Normal forceUsually
Fn = - Fgy
4. Friction
Ff =
f Fn
2. Break up Fg into its components Fgx and Fgy
Fgy = Fg cos
Fgx = Fg sin x
y
ExampleA 10 Kg block is resting on a 35o hill. Find the magnitude of all of the forces acting on the block.
ExampleA 10 Kg block is resting on a 35o hill. Find the magnitude of all of the forces acting on the block.
Fg = (10 Kg)(9.8 N/Kg)
Fn = -FgyFgy = -98 N cos 35 o
Fgx = 98 N sin 35o
Ff = -Fgx
x
y
ExampleA 10 Kg block is resting on a 35o hill. Find the magnitude of all of the forces acting on the block.
Fg = (10 Kg)(9.8 N/Kg)
Fn = -FgyFgy = -98 N cos 35 o
Fgx = 98 N sin 35o
Ff = -FgxFg = 98 N
Fgy = - 80.3 N
Fgx = 56.2 N
Fn = 80.3 N
Ff = -56.2N
Example
Fg = (10 Kg)(9.8 N/Kg)
Fn = -FgyFgy = -98 N cos 35 o
Fgx = 98 N sin 35o
Ff = -Fgx
Fgy = - 80.3 N
Fgx = 56.2 N
Fn = 80.3 N
Ff = -56.2N
All of the forces are balanced (F = 0 ) so this block will not accelerate.
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 = 3.0 Kg(9.8 N/Kg)
Fn = - Fgy
Fgy
Fgx
Fg = 29.4 N
Fgy = -22.5 N
Fgx = 18.9 N
Fn = - Fgy = 22.5 N
Fgx is unbalanced. So F = ma
Fgx = ma, 18.9 N = (3.0 Kg) aa = 18.9 N/ 3.0 Kg
a = 6.3 m/s2
x
y