Elevator and acceleration problem

In summary, a 60.0 kg student's apparent weight is 900.6N when she is accelerating upwards at 5.20 m/s2 while riding the elevator at the CN Tower in Toronto. When the elevator is moving downwards, her apparent weight would be 588.6N. The normal force, which represents her apparent weight, can be calculated by adding the gravitational force and the pseudo force resulting from the elevator's acceleration. This can be expressed using the equation Fn = ma + Fg. The unit of measurement for weight may vary depending on context, with some using kgf as a measure of force.
  • #1

Homework Statement


A 60.0 kg student is riding the elevator at the CN Tower in Toronto. Calculate her apparent weight if she is accelerating upwards at a rate of 5.20 m/s2. Calculate her apparent weight if she is riding down the elevator. Include a solution with the derived equation for each.

Homework Equations


F=ma

The Attempt at a Solution


I have drawn the free body diagram for the first part of the question and I think that the equation that I should use is Fn-Fg = ma. However I'm not sure how to get the normal force, do I substitute the numbers in and solve for it, or is there another way I get it? Any help would be much appreciated, I go to a self-directed school and there are no physics teachers here today for me to ask!
 
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  • #2
Looking at it again, I think I understand. Am I trying to find the normal force?
 
  • #3
Apparent weight is the normal force, yes.
 
  • #4
Ok! So now I have:
Fn = ma +fg
Fn = (60kg)(5.20m/s2) + (60kg)(9.81m/s2)
Fn = 312N + 588.6N
Fn = 900.6N
Then do I divide that by 9.81m/s2 to get it into kg? The answer would be 91.8 kg.
 
  • #5
pauladancer said:
Ok! So now I have:
Fn = ma +fg
Fn = (60kg)(5.20m/s2) + (60kg)(9.81m/s2)
Fn = 312N + 588.6N
Fn = 900.6N
Then do I divide that by 9.81m/s2 to get it into kg? The answer would be 91.8 kg.
That depends. Technically kg is the unit of mass and weight is measured in N. You'll have to ask whoever set the question what they meant. I would play safe and leave it in N.
 
  • #6
I see. I'll ask someone on Monday. Thank you so much for your help!
 
  • #7
pauladancer said:
I see. I'll ask someone on Monday. Thank you so much for your help!
Some people use kgf as a measure of force. 1 kgf is the force that gravity would impose on a mass of 1 kgm. To convert N to kgf, divide by 9.81.
 
  • #8
When we want to find the weight of a body, we can sort of imagine a weighing machine kept under the body. The weighing machine shows the Normal force exterted on it. So yes, we essentially need to find the normal force between the body and the contact. That would be given by the sum of the gravitational force [acts downward,obviously] and the pseudo force exerted due to the upward acc of the elevator [acts downward]
 

1. What is an elevator and acceleration problem?

An elevator and acceleration problem refers to a scenario where an elevator is either accelerating or decelerating while carrying passengers. This can result in a change in the perceived weight of the passengers due to the force acting upon them.

2. How does acceleration affect the weight of passengers in an elevator?

Acceleration affects the weight of passengers in an elevator by changing the normal force acting on them. As the elevator accelerates upwards, the normal force increases, making passengers feel heavier. Conversely, when the elevator accelerates downwards, the normal force decreases, making passengers feel lighter.

3. What is the difference between acceleration and velocity in relation to elevators?

Acceleration and velocity are both measures of the motion of an elevator, but they are distinct concepts. Acceleration refers to the rate of change of velocity, while velocity is the speed and direction of the elevator's motion. In an elevator and acceleration problem, the acceleration can change the velocity of the elevator, causing a change in the perceived weight of passengers.

4. How does the direction of acceleration affect the weight of passengers in an elevator?

The direction of acceleration directly affects the weight of passengers in an elevator. When the elevator is accelerating upwards, passengers will feel heavier, while accelerating downwards will make them feel lighter. If the elevator is at a constant velocity, there is no acceleration and passengers will feel their regular weight.

5. How can engineering principles be applied to solve elevator and acceleration problems?

Engineering principles, such as Newton's laws of motion and the principles of statics and dynamics, can be applied to solve elevator and acceleration problems. By analyzing the forces acting on the elevator and passengers, engineers can determine the acceleration and its effects on the weight of passengers. This knowledge can then be used to design and optimize elevators for safe and comfortable rides.

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