Download Frictionless Inclined Planes

GUIDED NOTES AND INTRODUCTORY PRACTICE FOR
INCLINED PLANES
Things I should already know:
Gravitational force is calculated by Fg=_______. Force weight is calculated by Fw=___________. The normal force arises
from ____________________________________________________, and always acts _____________________ to the
surface of contact. Most of the time, the numerical value of
FN is equal to ______.
In each of the diagrams below, draw in a vector correctly representing the normal force:
mg
mg
mg
mg
In the section below, highlight, underline and notate the important information… i.e. Take
Notes.
Inclined Planes
(source: http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section3.rhtml)
What we call wedges or slides in everyday language are called inclined planes in physics-speak. From
our experience on slides during recess in elementary school, sledding down hills in the winter, and
skiing, we know that when people are placed on slippery inclines, they slide down the slope. We also
know that slides can sometimes be sticky, so that when you are at the top of the incline, you need to
give yourself a push to overcome the force of static friction. As you descend a sticky slide, the force of
kinetic friction opposes your motion. In this section, we will consider problems involving inclined
planes both with and without friction. Since they’re simpler, we’ll begin with frictionless planes.
Frictionless Inclined Planes
Suppose you place a 10 kg box on a frictionless 30º inclined plane and
release your hold, allowing the box to slide to the ground, a horizontal
distance of d meters and a vertical distance of h meters.
Before we continue, let’s follow those three important preliminary steps for solving problems in
mechanics:
1. Ask yourself how the system will move: Because this is a frictionless plane, there is nothing to stop the box
from sliding down to the bottom. Experience suggests that the steeper the incline, the faster an object will
slide, so we can expect the acceleration and velocity of the box to be affected by the angle of the plane.
2. Choose a coordinate system: Because we’re interested in how the box slides along the inclined plane, we
would do better to orient our coordinate system to the slope of the plane. The x-axis runs parallel to the plane,
where downhill is the positive x direction, and the y-axis runs perpendicular to the plane, where up is the
positive y direction.
3. Draw free-body diagrams: The two forces acting on the box are the force of gravity, acting straight
downward, and the normal force, acting perpendicular to the inclined plane, along the y-axis. Because we’ve
oriented our coordinate system to the slope of the plane, we’ll have to resolve the vector for the gravitational
force, mg , into its x- andy-components. If you recall what we learned about vector decomposition in Chapter 1,
you’ll know you can break mg down into a vector of magnitude cos 30º in the negative y-direction and a vector
of magnitude sin 30º in the positive x-direction. The result is a free-body diagram that looks something like
this:
Decomposing the mg vector gives a total of three force vectors at work in this diagram: the ycomponent of the gravitational force and the normal force, which cancel out; and the x-component of
the gravitational force, which pulls the box down the slope. Note that the steeper the slope, the
greater the force pulling the box down the slope.
Note! There are some really important things to understand and get right about decomposing the
gravitational force vector.
1) the force weight, or Fg, always acts directly towards the center of the Earth, or straight
down
2) when drawing the right triangle showing the horizontal and vertical components of Fg,
make sure that the true Fg vector, that is straight down, is ALWAYS the hypotenuse of that right
triangle
Circle the correct formation:
mg
mg
3) The angle that the ramp surface makes with the floor is equivalent to the angle
between Fg and the vertical component.
Inclined Planes With Friction
There are two significant differences between frictionless inclined plane problems and inclined plane
problems where friction is a factor:
1. There’s an extra force to deal with. The force of friction will oppose the downhill component of the
gravitational force.
Notice that in the example above, the block still wants to slide down the ramp, so the friction
force directly opposes that by pointing up and being parallel to the surface of the ramp.
What would change about the friction force if we were applying a pulling force on the block to
accelerate it up the ramp?
Okay… let’s make sure we get it.
Look at the two diagrams below. Which box will slide down more quickly? ______________ Your everyday
life experience helps you quickly decide, now let’s prove it with our physics. The block has a 3kg mass. Ignore
friction for this example.
(A)
(B)
10°
45°
1) Draw in Fg and its components on both diagrams.
2) Calculate the x-component of each and compare. Which has a larger x-component? This numerically
shows why your above answer is correct.
x-component calculations:
A: (mg)sin10° = (3 kg * 9.81 m/s2)sin10° = 5.11 N
B: (mg)sin45° = (3 kg * 9.81 m/s2)sin45° = 20.8 N
Now, let’s practice with friction acting…
Look at the diagram below. If the coefficient of static friction is .523, will the 4.5 kg block slide down when
placed on the ramp?
30°
1) Draw in all of the forces: Fg, FN, and Ffr
2) Calculate the x-component of gravitational force, and calculate the force of static friction
- use (mg)sin30° and FN is equal and opposite to (mg)cos30° and Ffr = µFN
3) Then compare those two forces… if friction is a larger value than the x-component of gravitational
force, the block will not slide. And vice-versa, it will begin to slide.
Check your answers with the key!
Now work on the physicsclassroom.com “Inclined Plane Analysis” worksheet.
Check your answer with the key.
Finally, work the word problems and turn them in for a grade. They are due at the beginning of the next block.