Confused about F=MA problem

In summary, the reaction force of the lift upon the object would be equal to the object's weight (m*g) when the elevator is stationary. When the elevator is descending with an acceleration of 3m/s, the reaction force and weight both equal m*(-3 - (-9.8)), and when the elevator is decelerating at a rate of 3m/s, the reaction force and weight both equal m*(3 - (-9.8)). It is important to consider the direction of acceleration in relation to the gravitational force acting on the object. A good resource for understanding physics concepts at different levels is Khan Academy (www.khanacademy.org).
  • #1
joelio36
22
1
1. Question
A lift has a object of mass 5kg inside. Find the reaction force of the lift upon the object, and the object's relative weight when:
The lift is descending with an acceleration of 3m/s
The lift is descending with constant speed
The lift is decelerating at a rate if 3m/s (while descending)

2. My attempts and reasoning
I am confused about this question. I understand that on Earth the acceleration due to gravity is 9.8m/s/s, and the gravitational field is 9.8n/kg. So when the lift is at constant speed, acceleration is 0, but their is still an acceleration due to gravity (otherwise the object would lift off floor), so i assume weight & reaction = (9.8 * 5)= 49N
When the lift is descending at 3m/s/s, does this mean total acceleration is 9.8 + 3 = 12.8m/s, so weight and reaction both equal 12.8*5= 64N?
When the lift is retarding at 3m/s/s, total acceleration= 9.8-3= 6.8N, therefore weight and reaction both = 6.8 * 5= 34N.

I have no idea if my logic is right, close, or no where near, since this is what my buddy is doing, I haven't done it yet, because we are in different classes. are my answers right, if not, why?

PS- does anyone know a good website to help explain physics concepts at a-level/K10-12, undergrad level?
 
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  • #2
You have to consider how the forces are acting on the object in this situation.

When the elevator is stationary, there is a gravitational force equal to mg acting downwards. Since the object is stationary, the net force must be zero, thus the upward normal force has to equal the force of gravity (m*g).

When the elevator is accelerating downwards, the object has a total acceleration of 3m/s downward. Thsi means that the normal force is less than the gravitational force, and the net force on the object is Fn + mg = m*(-3). So Fn would be m * (-3 - (-9.8)) taking the downward direction as negative.

For the case when the elevator is decelerating, the same analysis applies, only now the acceleration is 3 m/s upward.
 
  • #3


First, it is important to clarify that the unit for acceleration is m/s^2, not m/s. So in the first scenario where the lift is descending with an acceleration of 3m/s^2, the total acceleration would be 9.8 + 3 = 12.8m/s^2, not 12.8m/s.

Your reasoning for the weight and reaction force being equal to 12.8 * 5 = 64N is correct. This is because the object is experiencing both the acceleration due to gravity (9.8m/s^2) and the additional acceleration from the lift (3m/s^2), resulting in a total acceleration of 12.8m/s^2. Therefore, the object's weight and the reaction force from the lift will both be equal to 64N.

In the second scenario where the lift is descending with constant speed, the object is still experiencing the acceleration due to gravity (9.8m/s^2) but there is no additional acceleration from the lift. This means that the object's weight and the reaction force from the lift will both be equal to 9.8 * 5 = 49N, as you correctly stated.

In the third scenario where the lift is decelerating at a rate of 3m/s^2, the total acceleration would be 9.8 - 3 = 6.8m/s^2. Your calculation for the weight and reaction force being equal to 6.8 * 5 = 34N is correct.

As for a good website for understanding physics concepts at different levels, Khan Academy and Physics Classroom are both great resources. Additionally, consulting your textbook or speaking with your teacher/professor can also be helpful in understanding and clarifying concepts.
 

1. What is the formula for F=MA?

The formula for F=MA is Force = Mass x Acceleration. This formula is used in Newton's second law of motion, which states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. In other words, the greater the mass of an object, the more force is needed to accelerate it.

2. How do I solve F=MA problems?

To solve F=MA problems, you need to know the values of force, mass, and acceleration. You can use the formula F=MA to calculate any one of these values if you know the other two. It is important to make sure that all units are consistent (e.g. if force is given in Newtons, mass should be in kilograms and acceleration in meters per second squared) and to pay attention to the direction of the force and acceleration vectors.

3. Can F=MA be used for all types of motion?

F=MA can be used for linear motion, where an object moves in a straight line, as well as rotational motion, where an object rotates around an axis. However, it is important to note that the formula may need to be modified for different types of motion, such as circular motion or motion on an inclined plane.

4. What is the significance of F=MA in physics?

F=MA is a fundamental formula in physics as it relates to one of Newton's laws of motion. It helps us understand the relationship between force, mass, and acceleration and how they affect the motion of objects. This formula is used in many fields, including engineering, mechanics, and astronomy, to make predictions and solve problems related to forces and motion.

5. Are there any limitations to using F=MA?

While F=MA is a useful formula, there are some limitations to its application. It assumes that the object is moving in a vacuum and that there are no external forces acting on it. In real-world scenarios, there are often other factors to consider, such as friction, air resistance, and other external forces, which may affect the motion of the object. In such cases, more complex equations and models may be needed to accurately describe the motion of the object.

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