Why does the apparent weight decrease when a body accelerates down a rope?

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When a body accelerates down a rope, its apparent weight decreases due to the relationship between gravitational force and the normal force exerted by the rope. Apparent weight is defined as the force felt by the body that counters gravity, which is lessened when the body accelerates downward. According to Newton's laws, when a net force acts on the body, it changes its state of motion, leading to a reduction in the normal force. If the body is in free fall, the normal force becomes zero, resulting in a sensation of weightlessness. Thus, the reduction in apparent weight occurs because the body is accelerating downward, not the other way around.
Kaushik
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When a body accelerates down(climbs down) a rope, why does its apparent weight reduce? Oh wait, does it even reduce in the first case?

If yes, is it because of the Newtons first law of motion, which states that a body in motion tends to be in motion and a body at rest tends to be at rest.

So when a body starts to accelerate downwards, it tends to be at rest, so there is a some random upward force that acts that tries to keep the body at the rest, which reduces its apparent weight.
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Kaushik said:
Summary: Why does the apparent weight of the body decrease while a body accelerates (climbs down) a rope?

When a body accelerates down(climbs down) a rope, why does its apparent weight reduce? Oh wait, does it even reduce in the first case?
Do you know what the definition of weight is ?

Kaushik said:
Summary: Why does the apparent weight of the body decrease while a body accelerates (climbs down) a rope?If yes, is it because of the Newtons first law of motion, which states that a body in motion tends to be in motion and a body at rest tends to be at rest.

does it ? ... have a look at this ...
https://en.wikipedia.org/wiki/Newton's_laws_of_motion
Kaushik said:
Summary: Why does the apparent weight of the body decrease while a body accelerates (climbs down) a rope?

So when a body starts to accelerate downwards, it tends to be at rest, so there is a some random upward force that acts that tries to keep the body at the rest, which reduces its apparent weight.

if it accelerating, how can it be at rest ?
 
Kaushik said:
Summary: Why does the apparent weight of the body decrease while a body accelerates (climbs down) a rope?

So when a body starts to accelerate downwards, it tends to be at rest, so there is a some random upward force that acts that tries to keep the body at the rest, which reduces its apparent weight.
There's two issues here. 1. What is the apparent weight? 2. Cause and effect.

1. What is the apparent weight.
Your real weight is the force of gravity on your body. There's no way to change that, apart from getting away from the Earth with a spaceship. Apparent weight is the force on your body that stops it from falling. If you are standing on the ground that will be the normal force from the ground on your feet, If you're hanging from a rope that will be the friction force from the rope on your hands. Of course if you don't accelerate, this force must be equal and in opposite direction from the force of gravity.
Apparent weight is the force that you actually feel.

2. Cause and effect.
Acceleration does not cause a reduction of the apparent weight. You cause acceleration by reducing your apparent weight. The easiest way is to let go of the rope. Your apparent weight will be 0, and because the force of gravity is still there, you will rapidly accelerate downwards. If you climb down the rope, you will just reduce the apparent weight for a short period.

If course if you want to go down, you must reduce your apparent weight, and if you see someone accelerating downwards, you can calculate that his apparent weight must be smaller, but neither of these dmean that acceleration is the cause of the reduction of the apparent weight.
 
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davenn said:
if it accelerating, how can it be at rest ?
It tries.
 
davenn said:
Do you know what the definition of weight is ?
Weight is the force acting on a body due to gravity.
 
willem2 said:
Apparent weight is the force that you actually feel.
Is it the net vertical force acting on the body in the figure attached above?
 
willem2 said:
Acceleration does not cause a reduction of the apparent weight. You cause acceleration by reducing your apparent weight.
Oh yes, it is the force which cause the acceleration, not the other way around.
 
davenn said:
does it ? ... have a look at this ...
So first law states that, a body that is in motion or at rest will not be changed until ## F_{net} ≠ 0 ##. When there is some net force acting on it the body changes its state of motion(It starts to accelerate).
Am I right?
So, is the reduction of apparent weight somehow related to this?

Thanks.
 
Kaushik said:
Weight is the force acting on a body due to gravity.
Once you have subtracted out all the other forces on a body, "weight" is the remaining force that is required to explain the body's acceleration or lack thereof.

When defined in this way, "weight" depends crucially on what standard you measure your acceleration against. Change the standard for being at rest and you change your "weight".
 
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Imagine the body standing on a scale.

First case:
Your body is not accelerating downward. You are in equilibrium. The normal force (n) exerted by the scale is equal and opposite to the weight (mg) of the body. Hence,

F = ma = n - mg = 0

Second case:
The body is accelerating downward in the negative direction:

F = ma = n - mg < 0

Hence

n < mg

The normal force (n) is your apparent weight and is now less than your actual weight.

The reason it has decreased is when talking about apparent weight we are talking about the weight of your body against the scale or conversely the scale against your body which is the normal force. If the scale is freely falling along with your body it exerts no normal force and so your weightless. But if it is not in free fall, that is the acceleration is less than g, then it must be preventing you from free falling since it is supporting you. But you are still accelerating downward a little bit. So your apparent weight is somewhere between 0 and your weight mg.
 
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