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ELEVATOR PHYSICS

ELEVATOR PHYSICS.

alfreda-hughes
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ELEVATOR PHYSICS

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  1. ELEVATOR PHYSICS

  2. 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 ?

  3. (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

  4. 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

  5. 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 = - FW FN = -(25kg)(-9.8 m/s2) FN = 245 N FN FA 25 kg FW

  6. 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

  7. 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?

  8. Given: m = 20 kg FA = 50 N at 30 o from horizontal Find a and FN FA FN FAy 20 kg 30 o 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 FAx FW

  9. FA FN a = FAx= 43.3 N = 2.17 m/s 2 m 20 kg FAy 20 kg 30 o FAx FW 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

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