Post on 31-Dec-2015
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A passenger of mass m= 72.2 kg A passenger of mass m= 72.2 kg stands on a bathroom scale in an stands on a bathroom scale in an elevator. What are the scale readings elevator. What are the scale readings when the cab is stationary, when it is when the cab is stationary, when it is moving up and moving down?moving up and moving down?(a) Find the (a) Find the general equationgeneral equation for the for the scale reading, whatever the vertical scale reading, whatever the vertical motion of the cab.motion of the cab.(b) What does the scale read if the (b) What does the scale read if the cab is stationary or moving upward cab is stationary or moving upward at a constant 0.50 m/s?at a constant 0.50 m/s?(c) What does the scale read if the (c) What does the scale read if the cab accelerates upward at 3.20 m/scab accelerates upward at 3.20 m/s22 and downward at 3.20 m/sand downward at 3.20 m/s22 ? ?
(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 NWhen the acceleration is directed downward, Fapp = (72.2 kg) (-3.2m/s2 - -9.8m/s2) = 477 N
DESIGNATION OF FORCES DESIGNATION OF FORCES IN A FREE BODY DIAGRAMIN A FREE BODY DIAGRAM
FFAA = applied force = applied force
FFNN = normal force; Force which is = normal force; Force which is perpendicular to the surfaces in contactperpendicular to the surfaces in contact
FFWW = weight of the object; this is always = weight of the object; this is always directed downwarddirected downward
FFTT = force as it is acting on a rope or a = force as it is acting on a rope or a wirewire
Neglecting friction, determine the normal force Neglecting friction, determine the normal force and horizontal force needed to accelerate a 25 and horizontal force needed to accelerate a 25 kg grocery cart from rest to a velocity of 0.45 kg grocery cart from rest to a velocity of 0.45 m/s in 1.35 s.m/s in 1.35 s.
Given:Given: m= 25 kgm= 25 kg Vix = 0Vix = 0
Vfx = 0.45 m/sVfx = 0.45 m/s t = 1.35 st = 1.35 s
Find Find FFNN and and FFAA
25 kg
The cart has no vertical motion since there is no unbalanced force along the y-axis.
FW + FN = 0
FN = - FW
FN = -(25kg)(-9.8 m/s2)
FN = 245 N FW
FN FA
To solve for the acceleration:To solve for the acceleration:
a = a = Vfx-Vix Vfx-Vix = = 0.45 m/s – 0 m/s0.45 m/s – 0 m/s = = 0.33 m/s0.33 m/s22
t 1.35 s t 1.35 s
To solve for the applied force, FTo solve for the applied force, FAA::
FFAA = = mm a = a = (25 kg)(25 kg) (0.33 m/s (0.33 m/s22 ) )
= = 8.25 N 8.25 N
A 20-kg bag is being pulled along A 20-kg bag is being pulled along the floor by a tourist with a force the floor by a tourist with a force of 50 N. The force is applied on of 50 N. The force is applied on the handle that forms an angle of the handle that forms an angle of 3030oo with the horizontal. What is with the horizontal. What is the acceleration of the bag? How the acceleration of the bag? How much force is exerted by the floor much force is exerted by the floor on the bag if friction is on the bag if friction is neglected?neglected?
Given: m = 20 kgGiven: m = 20 kg
FFAA = 50 N at 30 = 50 N at 30 oo from horizontal from horizontal
Find Find aa and and FFNN
Solve for the FAx and FAy
FAx = FA cos θ
= (50 N) (cos 30 o) = 43.3 N
FAy = FA sin θ
= (50 N) (sin 30 o) = 25 N
20 kg 30 o
FA
FAy
FAx
FN
FW
20 kg 30 o
FA
FAx
FN
FW
FAya = FAx = 43.3 N = 2.17 m/s 2
m 20 kg
All forces along the y-axis:
FN + FAy + FW = 0
FN = - FW – FAy
= - (20kg)(-9.8 m/s2) – 25 N
= 196 N – 25 N = 171 N