ch 6 notes - 6.1 Equilibrium -(F(net)(x)=sigma(x)(F(i)(x)=0

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
6.1 Equilibrium -(F(net))(x)=sigma(x)(F(i))(x)=0 -(F(net))(y)=sigma(y)(F(i))(y)=0 -the equilibrium condition of Equations 6.1 applies only to particles, which cannot rotate. -equilibrium of an extended object, which can rotate, requires an additional condition 6.2 Using Newton's Second Law -the essence of Newtonian mechanics can be expressed in two steps: 1. forces on an object determine its acceleration a=F(net)/m 2. object's trajectory can be determined by using a in the equations of kinematics -(F(net))(x)=sigma(F(x))=m*a(x) -(F(net))(y)=sigma(F(y))=m*a(y) 6.3 Mass, Weight, and Gravity -mass is an intrinsic property of an object -a pan balance would give the same result on another planet because it does not depend on the strength of gravity -gravity is an attractive, long-range force between any two objects -Newton's law of gravity: F(1 on 2)=F(2 on 1)=Gm(1)m(2)/r^2 -G=6.67*10^-11 Nm^2/kg^2, called the gravitational constant -F(g)=F(planet on m)=(GMm/R^2, straight down) -For motions whose vertical and horizontal extents are less than roughly 10 km, we can approximate the earth as a flat surface that pulls on objects with a gravitational force
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

ch 6 notes - 6.1 Equilibrium -(F(net)(x)=sigma(x)(F(i)(x)=0

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online