Chapter 7.1-7.2 Notes
Steps for solving ANY 2-d force problem:
A. Draw a free body diagram.
B. List all forces in the x direction
C. List all forces in the y direction
D. List all forces that are not in the x or y direction
E. Draw a triangle for each force in step D with the x and y components.
F. Break down all of the forces from step D into their x and y direction components.
G. Write the net force equation for the x direction (the sum of all the x direction forces).
H. Write the net force equation for the y . direction (the sum of all the y direction forces).
I. Determine the acceleration in the x direction.
J. Determine the acceleration in the y direction.
K. Determine the net force using F net=ma for the x direction.
L. Determine the net force using F net=ma for the y direction.
Click here for help on determining the coefficient of friction for a block sliding down the ramp at a constant velocity.
7.1: Equilibrium and the Equilibrant
An object is in equilibrium when the net force on it is zero. Net force may be the resultant of 2 or more forces acting concurrently, but the sum of all these forces is zero.
Such a condition means that the object is either at rest or in motion with a constant velocity (acceleration = zero) .
If the sum of all the forces acting concurrently is not qero, then the object is not in equilibrium
If one more force can be added that will bring the object into equilibrium, such a force is called the equilibrant.The equilibrant is equal in magnitude to the non-zero resultant of all other forces, but opposite in direction.
Gravitational Force and Inclined Planes
Finding FN, the normal force.
Instead, we resolve the weight x and y components. Fg y is constructed perpendicular to the frictional surface. Its magnitude can be calculated when the weight mg and angle are known.
However, due to the geometry of the diagram and the fact we need the angle from the positive x axis, which is tilted, we must subtract this angle from 90 degrees for the remainder of this problem.
The normal force FN is equal in magnitude to the y component of gravity:
Finding the Ff, the frictional force.
The Frictional force Ff is equal in magnitude to Fgx.
7.2 Projectile motion is a particular kind of 2 dimensional motion.
For most of the problems we will deal with in this section, we make the following assumptions:
Independence of Motion in Two Dimensions
Imagine a bullet fired from a rifle barrel that is horizontal to the ground, as in the diagram below . Then imagine a second bullet that is dropped directly to the ground, from the same heigth as the rifle barrel, and at exactly the same time the first bullet is fired. Which bullet will reach the ground first? Answer... barring obstacles and air resistance, they will both reach the ground at the same time! Although the "fired" bullet is moving horizontally with more or less constant velocity, it is accelerating toward the ground under the same rules of gravity as the bullet that was just dropped, (-9.8 m/s/s).
|Look at this apparatus. Above, you can see the spring loaded plunger with a resting position on the left, and an extension on the right. When two spheres are loaded onto the apparatus as shown, the spring plunger strikes the left sphere, giving it a horizontal velocity, while the simultaneously dropping the right sphere straight down.|
|At right is a short movie showing the apparatus above in action. You will need the Quicktime plugin for your browser. When the movie is loaded, use the player controls to still-step the movie. You will notice the left sphere follows a parabolic pathway to the ground when it is fired, while the right one goes straight down. Both spheres start at exactly 0.95 meters above the floor, and even though one has some horizontal velocity, they both strike the floor at the same time!|
|Procedure for Solving Projectile Motion Problems: When analyzing motion in two dimensions, separate the motion into its vertical component, and its horizontal component .
In projectile motion problems like the one above, it is handy to set up your problem solving structure as follows:
Use the same method outlined in Chapter 3 to solve problems in this section.
|The image at the right shows the ball from the apparatus described above just as it hits the ground. It arrived from the right on it's parabolic flight, and strikes the floor at precisely 0.69 meters horizontally from it's starting position. It's original position was 0.95 meters above the floor.
Calculate the time required for the ball to reach the ground, and the horizontal velocity:
|Here are some notes about how to find Vi when you only know range and launch angle.
BIG TIME HINT: Be sure that when you use any of the motion formulas listed here that all velocities and distances are in the same dimension within the formula. That is, don't use a horizontal distance in the same formula that includes a vertical acceleration!
Here's a .pdf that summarizes formulas, symbols and values used in parabolic motion
Here's a cool on-line page that does all kinds of trajectory calculations, it could help with your homework!