Page 75 - Schaum's Outline of Theory and Problems of Applied Physics

60 LAWS OF MOTION [CHAP. 5

reaction force to F:
w app = mg + ma
Apparent weight = Actual weight + accelerating force

Fig. 5-7
If a is positive, corresponding to an upward acceleration, w app > mg, and the scale reading will be greater than
the person’s 700-N actual weight. If a is negative, corresponding to a downward acceleration, w app < mg, and
the scale reading will be less than 700 N. If the cable that supports the elevator breaks and the elevator falls
freely, a =−g, and w app = mg − mg = 0; the woman is “weightless.” When the elevator is at rest or moving
at constant speed up or down, a = 0 and w app = mg.

SOLVED PROBLEM 5.23

The scale in the elevator of Fig. 5-7 reads 600 N. Find the elevator’s acceleration.
Here w app < mg, so the elevator’s acceleration is downward. The person’s mass is
mg 700 N
m = = 2 = 71.4kg
g 9.8 m/s
From w app = mg + ma we have
w app 600 N 2 2 2
a = − g = − 9.8 m/s = (8.4 − 9.8) m/s =−1.4 m/s
m 71.4kg

TWO AND THREE DIMENSIONS
When the forces that act on a body do not all lie along the same straight line, vector methods are needed to
analyze the situation. In component form, the second law of motion is

F x = ma x
F y = ma y
F z = ma z
where F = F x + F y + F z is the net force on a body and a = a x + a y + a z is its acceleration. The best way to
orient the coordinate axes depends on the problem. Often forces involved lie in a plane, in which case only two
force components and two acceleration components are needed.

SOLVED PROBLEM 5.24

◦
A 60-kg sprinter presses down on the ground with a force of 1000 N at an angle of 30 with the horizontal
at the start of a race. What is his forward acceleration as his legs straighten out?