ge 253 week 4 lecture slides

8/8/2019 GE 253 Week 4 Lecture Slides

1/11

Instructor: Manish Narayan

Physics


2/11

First Law: An object at rest remains at rest and an object in

constant motion remains in constant motion unlessacted upon by a net external Force.

Second Law: F = ma (Force equals mass times acceleration) Or acceleration is directly proportional to the Force but

inversely proportional to the mass.

Third Law: For every action force there is an equal and opposite

reaction force.


3/11

F = ma

Force = mass x acceleration

= kg x m/s

2

or 1 N 1 N = 1 Newton

One Newton of force is required to cause a 1 kgobject to accelerate at 1 m/s2.

1 kN = 1000 N


4/11

If the net force on an object is 0 (balanced) , thenthe object is said to be in equilibrium (meaning atrest or at a constant velocity)

If the net force on an object is not zero, then the

object must be accelerating:

If I apply a force to the right of 100 N and anotherforce in the opposite direction of 50 N then the netforce on the box is: -50 + 100 = 50 N (The box willaccelerate to the right!)

100 N-50 N


5/11

The Earth pulls us down with a gravitationalforce and we call this our Weight (W)

Using Newtons second law and the fact thatEarths gravitational acceleration is 9.8 m/s2:

W=mg

Weight is also measured in Newtons (N) as it is

a Force. My weight is W= (67 kg)(9.8 m/s2) = 657 N on

Earth.


6/11

When an object is in contact with a surface (i.e., table orwall) then there is a perpendicular force the surface exertson the object which is known as normal force:

Weight is the force acting down due to gravity.

The contact or normal force is the table pusing up on thegreen box (N)

If you apply a force to the left (F) then the frictional forcewill be in the opposite direction of motion (Ff)

Frictional force will be discussed in next slide.


7/11

The frictional force on an object sliding on asurface is:

mu () is called the frictional coefficient and N isthe normal force.

There are two types of frictional coefficients: Coefficient of Static Friction: s Coefficient of Kinetic Friction: k Initially it is hard to push an object since one must

overcome static friction, but once it gets movingthen a smaller force is needed to keep it moving(kinetic friction).

NFf Q!


8/11

Remember a vector is a quantity that has bothmagnitude and direction (velocity, acceleration,force, etc.)

Vectors can be resolved into components:

F

Fy

Fx

U

22

sin

cos

yx

y

x

FFF

FF

FF

!

!

!

U

U


9/11

Assume a Force F=100 N is applied at an angleof 30 degrees in the previous diagram:

First find the x component of Force:

Next find the y component of Force:

NewtonsF

FF

x

x

6.8630cos100

cos

!!

! U

Ne t sx

x

0.5030sin00

sin

!!

! U


10/11

If there are many forces acting on an object withdifferent magnitudes and different directions thenjust break each force into components and finallyfind the net force in both the x & y directions:

Once the sum of the forces in each direction arefound then just use Pythagorean theorem to findthe resultant forces magnitude.

Finally use the tan-1 of the y component of force

divided by the x component of force in order tofind the direction angle of the resultant.

Example of resolving multiple forces on nextslide


11/11

The force in the left direction (-50 N) only has a x component of force:

Fx=-50 N The force (100 N) at an angle of 30 degrees has both x and y components

of force: Fx=100cos(30) = 86.6 N Fy=100sin(30) = 50.0 N Now sum all the forces in the x and y direction: F

x= -50 + 86.6 = 16.6 N

Fy= 50 N (only one component in y direction) Use Pythagorean Theorem to find resultant force:

The direction of the resultant force is :

-50 NQ

30!U

NewtonsF 7.52506.16 22 !!

Q6.7

6.6

50tan !

! U

ge 253 week 4 lecture slides

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