Download
objects on inclined planes n.
Skip this Video
Loading SlideShow in 5 Seconds..
Objects on Inclined Planes PowerPoint Presentation
Download Presentation
Objects on Inclined Planes

Objects on Inclined Planes

180 Vues Download Presentation
Télécharger la présentation

Objects on Inclined Planes

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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

  2. y y x x Inclined Plane Object on an inclined plane 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.

  3. y x Inclined Plane Rotate the axis This way the motion will be in the “X” and we are back to 1-D physics!

  4. Draw your free body diagram. Fn Ff y x Fg 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 5. Other forces?

  5. Get the values. y x Fgy = Fg cos q Fgx = Fg sin q Fn = - Fgy Ff = mf Fn • Force due to Gravity 2. Break up Fg into its components Fgx and Fgy 3. Normal force Usually 4. Friction Fg= m (9.8 N/Kg)

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

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

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

  9. Example Fgy = - 80.3 N Fgx = 56.2 N Fn = 80.3 N Ff = -56.2N Ff = -Fgx Fn = -Fgy All of the forces are balanced (SF = 0 ) so this block will not accelerate. Fgy = -98 N cos 35o Fgx = 98 N sin 35o Fg = (10 Kg)(9.8 N/Kg)

  10. 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. y Fn = - Fgy x Fgy Fgx Fg = 3.0 Kg(9.8 N/Kg) 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