Objects on Inclined Planes

I guess you could call them ramps if you want.

Inclined PlaneObject on an inclined plane

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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

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y