Forces on Inclined Planes Unit 3, Presentation 3

Inclined Planes  Inclined planes create unusual complications because the standard x-y plane is no longer the only direction of motion  The x-y plane must be rotated to most easily solve the problems.

An Inclined Plane Example Problem  Calculate the time it takes for the following block to slide down the inclined plane: 5 kg 40° 10 m  k =0.30 Assume that the block starts from rest.

Inclined Plane Example Problem First, draw a free body diagram for the block: mg FnFn frictionNote that, right now, only one of our forces is in the standard x-y plane (mg). If we re-orient the x-y plane with the positive x-axis along friction and the positive y-axis along the normal force, then we only have to break mg down into components: x y xy

Inclined Plane Example Problem Now, lets consider the forces in the x and y directions: x - directiony - direction Friction (negative) x-component of weight: Normal Force y-component of weight:

Inclined Plane Example Problem Now, lets set up separate equations for the x and y directions using Newton’s 2 nd Law: x - directiony - direction Using substitution Lets solve for acceleration in the x-direction and use that to find time using the kinematic equations.

Inclined Plane Example Problem Now we need to find the distance traveled using trigonometry: 10 m 40° x

Inclined Plane Example Problem Now, use the big kinematic equation to find time.

Another Inclined Plane Example Problem Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block begins to slip down?  2.50 kg  s =0.350

Another Inclined Plane Example First, draw a free body diagram for the block: mg FnFn friction x-direction y-direction Friction (negative) x-component of weight Normal Force y-component of weight

Another Inclined Plane Example Now, use Newton’s Second Law in both directions: x-direction y-direction When it just begins to move, acceleration just begins to increase above zero. To find the critical point, set a=0. Use Substitution Note that both m and g cancelled out of the problem!