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A passenger of mass m= 72.2 kg stands on a
bathroom scale in an elevator. What are the scale
readings when the cab is stationary, when it is
moving up and moving down?
(a) Find the general equation for the scale
reading, whatever the vertical motion of the cab.
(b) What does the scale read if the cab is
stationary or moving upward at a constant 0.50
m/s?
(c) What does the scale read if the cab accelerates
upward at 3.20 m/s2 and downward at 3.20 m/s2 ?
(a) To determine the apparent weight (Fapp), first we
find all of the forces in the y direction:
Ʃ Fy = Fapp + FW
Then we use Newton's Second Law.
may = Fapp + mg
Then solve for F
Fapp = may – mg = m (ay-g)
This works no matter what the acceleration is for the cab.
(b) When the acceleration is zero,
Fapp = (72.2 kg) (0 - -9.8m/s2) = 708 N
This is just the weight of the passenger when the
elevator is at rest or moving with a constant velocity.
(c) When the acceleration is directed upward,
Fapp = (72.2 kg) (3.2m/s2 - -9.8m/s2) = 939 N
When the acceleration is directed downward,
Fapp = (72.2 kg) (-3.2m/s2 - -9.8m/s2) = 477 N
DESIGNATION OF FORCES IN A
FREE BODY DIAGRAM
FA = applied force
FN = normal force; Force which is perpendicular
to the surfaces in contact
FW = weight of the object; this is always directed
downward
FT = force as it is acting on a rope or a wire
Neglecting friction, determine the normal force and
horizontal force needed to accelerate a 25 kg grocery
cart from rest to a velocity of 0.45 m/s in 1.35 s.
Given:
m= 25 kg
Vix = 0
Vfx = 0.45 m/s
t = 1.35 s
Find FN and FA
The cart has no vertical motion since there is no
unbalanced force along the y-axis.
FW + FN = 0
FN
FN = - FW
FN = -(25kg)(-9.8
FN = 245 N
FA
25 kg
m/s2)
FW
To solve for the acceleration:
a = Vfx-Vix = 0.45 m/s – 0 m/s = 0.33 m/s2
t
1.35 s
To solve for the applied force, FA:
FA = m a = (25 kg) (0.33 m/s2 )
= 8.25 N
A 20-kg bag is being pulled along the floor by
a tourist with a force of 50 N. The force is
applied on the handle that forms an angle of
30o with the horizontal. What is the
acceleration of the bag? How much force is
exerted by the floor on the bag if friction is
neglected?
Given: m = 20 kg
FA = 50 N at 30 o from horizontal
Find a and FN
FA
FN
20 kg
Solve for the FAx and FAy
FAx = FA cos θ
FAx
FW
= (50 N) (cos 30 o) = 43.3 N
FAy = FA sin θ
= (50 N) (sin 30 o) = 25 N
30
FAy
o
FA
FN
a = FAx = 43.3 N = 2.17 m/s 2
m
20 kg
20 kg
All forces along the y-axis:
FAx
FW
FN + FAy + FW = 0
FN = - FW – FAy
= - (20kg)(-9.8 m/s2) – 25 N
= 196 N – 25 N = 171 N
30
FAy
o