Flashcards in Newtonian Mechanics Deck (25):

1

## Force

###
Vector quantity describing the push/ pull on an object with units: Newton

F = ma

2

## Newton

### N = kg m/ s^2

3

## Mass

### Scalar quantity that measures an object's inertia.

4

## Weight

### A force that involves the gravitational pull of an object. W = mg

5

## Center of Gravity

### The point where the entire force can be thought to be applied that is usually the geometric center for a homogenous body.

6

## Newton's First Law

### A body at rest or in motion with constant velocity will remain that way unless a net force acts upon it

7

## Newton's Second Law

###
Σ F = ma

The forces can also be broken up into its components

When Σ F = 0, then a = 0

8

## Newton's Third Law

### Fa = - Fb

9

## Why would it be safer to collide with bales of hay than with a solid fence?

### It takes more time for a vehicle to slow down when colliding with a bale of hay, leading to a smaller average impact force.

10

## Normal Force

### The perpendicular component of a force, where N = mg on a flat plane and N = mg sin θ on an incline.

11

## Friction Force

### Antagonistic force dependent upon the objects involved, where F = μN

12

## How do you calculate the forces acting upon a block sliding on a frictionless incline?

###
1. Draw a free-body diagram (the block feels two forces: gravity and the normal force)

2. Assign x and y axes (preferably make x parallel to the incline)

3. Find the components of the forces (Wx = mg sin θ and Wy = mg cos θ, because the angle of incline equals the angle between the gravitational force and the normal force)

4. Only the forces in the x-direction affect the motion of the block, since there is no acceleration in the y direction (recall that x is parallel to the incline) [ΣFx = Wx = max and ΣFy = N - Wy = may = 0] {apply Newton's second law separately to each direction}

5. The length of the incline can be found with d = v0t + (at^2)/2

6. The vertical height of the incline can be found with sin θ = h/d

13

## Gravity

###
The attractive force felt by all forms of matter: F = Gm1m2/ r^2

G = 6.67 X 10^-11 N m^2/kg^2

r = distance between the centers of the masses

14

## Equilibrium

### When ΣF = 0

15

## Translational Equilibrium

### ΣFx = 0 and ΣFy = 0

16

## How do you calculate the forces acting upon a mass being held by a number of strings in equilibrium?

###
1. Draw a free-body diagram at the point of intersection, do not forget to include weight. The tension in the vertical component should be equal to the weight of the mass, because ΣFx = 0 and ΣFy = 0.

2. ΣFx = 0 = ΣT cos θ

3. ΣFy = 0 = ΣT sin θ - W

4. Substitute into the equations as necessary

17

## Rotational Equilibrium

### Rotational equilibrium depends on force magnitude, force direction, and the distance from the force to the axis of rotation. The greater the distance, the greater the change in rotational motion. Clockwise is negative and anti-clockwise is positive: Στ = 0

18

## Torque

### τ = rF sin θ, with SI units N m

19

## How do you calculate the forces with a balanced see-saw?

###
1. Take the point of the fulcrum as the pivot point to eliminate the normal force and weight of the see-saw from the equation because r = 0.

2. Στ = 0 = Σmgd, because θ = 90 on a perpendicular see-saw

3. N exerted by the fulcrum is equal to the sum of all weights acting on the see-saw

20

## Translational Motion

### Motion where there is no rotation

21

## Static Friction

### The force that must be overcome to set an object in motion.

22

## How is the magnitude of the acceleration for blocks connected by a string related?

### The magnitude of accleration for the blocks should be the same

23

## Centripetal Acceleration

### a = v^2/ r, pointed towards the center of motion

24

## What assumptions can be made when working with uniform circular motion?

### Object speed is constant and there is no tangential acceleration.

25