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Sliva [168]
4 years ago
4

A simple elevator ride can teach you quite a bit about the normal force as this rider below can (hopefully) tell you. There are

three different scenarios given, detailing the rider\'s experience in an unnamed hotel. For each scenario, calculate the normal force, FN,1-3, acting on the rider if his mass is m = 76.6 kg and the acceleration due to gravity g = 9.81 m/s2. In scenario 1, the elevator has constant velocity. In scenario 2 the elevator is moving with upward acceleration a2 = 4.84 m/s2. Finally, in scenario 3, unfortunately for the rider, the cable breaks and the elevator accelerates downward at a3 = 9.81 m/s2.
FN1= _______
FN2=_______
FN3=___________
Physics
1 answer:
Alina [70]4 years ago
6 0

Answer:

FN1 = 751.5 N  

FN2 = 1122.2 N  

FN3 = 0

Explanation:

Scenario 1 :

• The elevator has constant velocity.

  • The normal force, can adopt any value, as needed by Newton's 2nd Law, in order to fit this general expression:
  • Fnet = m*a
  • In the first scenario, as the elevator is moving at a constant speed, this means that no external net force is present.
  • The two forces that act on the rider, are gravity (always present, downward) and the normal force, as follows:
  • Fnet = Fn - m*g = m*a  (taking the upward direction as positive)

For scenario 1:

Fnet = 0 ⇒ Fn = m*g = 76.6 kg * 9.81 m/s² = 751. 5 N

Scenario 2:

  • In this scenario, the elevator has an upward acceleration of 4.84 m/s², so the Newton's 2nd Law is as follows:
  • Fnet = FN - m*g = m*a  
  • ⇒ FN = m* ( g+ a) = 76.6 kg* (9.81 m/s² + 4.84 m/s²) = 1,122.2 N

Scenario 3 :

  • As the elevator is in free fall, this means that a = -g, so, in this condition, the normal force is just zero, as it can be seen from the following equation:
  • FN-mg = m*a  
  • If a = -g,  
  • ⇒ FN -mg = -mg ⇒ FN=0

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4 years ago
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Answer:

R = (- 3.72î + 8.29j)

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Let î and j represent unit vectors along the x and y axis respectively.

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Let the x and y components of this vector be Aₓ and Aᵧ

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The components given magnitude and direction from the +x-axis are calculated as

Aₓ = A cos θ and Aᵧ = A sin θ

Aₓ = (8.78 cos 37°) = 7.01 m

Aᵧ = (8.78 sin 37°) = 5.28 m

A = (7.01î + 5.28j) m

Vector B has magnitude 8.26 m and direction 135° from the +x-axis

B = (Bₓî + Bᵧj) m

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