Objects on Inclined Planes I guess you could call them ramps if you want.

Inclined Plane Object 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. Friction Ff is always against the motion Ff 1. 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 force Usually Fn = - Fgy 4. Friction Ff =  f Fn 2. Break up Fg into its components Fg x and Fg y Fg y = Fg cos  Fgx = Fg sin  x y

Example A 10 Kg block is resting on a 35 o hill. Find the magnitude of all of the forces acting on the block.

Example A 10 Kg block is resting on a 35 o hill. Find the magnitude of all of the forces acting on the block. Fg = (10 Kg)(9.8 N/Kg) Fn = -Fgy Fgy = -98 N cos 35 o Fgx = 98 N sin 35 o Ff = -Fgx x y

Example A 10 Kg block is resting on a 35 o hill. Find the magnitude of all of the forces acting on the block. Fg = (10 Kg)(9.8 N/Kg) Fn = -Fgy Fgy = -98 N cos 35 o Fgx = 98 N sin 35 o Ff = -Fgx Fg = 98 N Fgy = N Fgx = 56.2 N Fn = 80.3 N Ff = -56.2N

Example Fg = (10 Kg)(9.8 N/Kg) Fn = -Fgy Fgy = -98 N cos 35 o Fgx = 98 N sin 35 o Ff = -Fgx Fgy = 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 40 o 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 = N Fgx = 18.9 N Fn = - Fgy = 22.5 N Fgx is unbalanced. So  F = ma Fgx = ma, 18.9 N = (3.0 Kg) a a = 18.9 N/ 3.0 Kg a = 6.3 m/s 2 x y