position, displacement, velocity, and acceleration (Sections 3.1–3.3)
Newton's second law (Sections 3.3 and 3.4)
addition and subtraction of vectors (Sections 2.2, 2.3, and 3.1)
vector components (Section 2.3)
internal and external forces (Section 2.5)
Mastering the Concepts
Essential relationships for constant acceleration problems: If ax is constant during the entire time interval Dt from ti until a later time tf = ti + Dt,
These same relationships hold for the y-components of the position, velocity, and acceleration if ay is constant.
The only force acting on an object in free fall is gravity. On Earth, free fall is
an idealization since there is always some air resistance. An object in free fall has an acceleration equal to the local value of the gravitational field <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0073512141/663812/ch4_2.JPG','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a>
For a projectile or any object moving with constant acceleration in the ±y-direction, the motion in the x- and y-directions can be treated separately. Since ax = 0, vx is constant. Thus, the motion is a superposition of constant velocity motion in the x-direction and constant acceleration motion in the y-direction.
The x- and y-axes are chosen to make the problem easiest to solve. Any choice is valid as long as the two are perpendicular. In an equilibrium problem, choose x- and y-axes so that the fewest number of force vectors have to be resolved into both x- and y- components. In a nonequilibrium problem, if the direction of the acceleration is known, choose x- and y-axes so that the acceleration vector is parallel to one of the axes.
Problems involving Newton's second law—whether equilibrium or nonequilibrium—can be solved by treating the x- and y-components of the forces and the acceleration separately.
An object that is accelerating has an apparent weight that differs from its true weight. The apparent weight is equal to the normal force exerted by a supporting surface with the same acceleration. A helpful trick is to think of the apparent weight as the reading of a bathroom scale that supports the object.
