The Lift Question / reactive force on an object in a moving lift

[druuuuuuuunnk]

Homework Statement

A mass of 80 kg rests on the floor of a lift. (Take g = 10ms-2). Calculate:

(a) the reaction force exerted on the object by the floor, when the lift is
travelling at a steady speed of 2 ms-1 downwards.

(b) the reactive force exerted on the object by the floor of the lift, as its
downward motion is brought to rest by a retardation of 4ms-2.

(c) the reactive force exerted on the object by the floor, when the lift is at rest.

Gravity = 10 m/s-2

Homework Equations

F=M.A

The Attempt at a Solution

(a) becuase there is a steady speed downwards, do i just use gravity as the accelleration and nothing else? so F= (80) x (10) F=800 N

/or/ is it F= (80) x (10+2) F=960 N

(b) F= (80) x (10-4) so F= 480 N

(c) F= (80) x (10) F=800

My teacher this year has been pretty useless if you could help me understand this i would be really gratful.

 

[Modulated]

Homework Statement

A mass of 80 kg rests on the floor of a lift. (Take g = 10ms-2). Calculate:

(a) the reaction force exerted on the object by the floor, when the lift is
travelling at a steady speed of 2 ms-1 downwards.

(b) the reactive force exerted on the object by the floor of the lift, as its
downward motion is brought to rest by a retardation of 4ms-2.

(c) the reactive force exerted on the object by the floor, when the lift is at rest.

Gravity = 10 m/s-2

Homework Equations

F=M.A

The Attempt at a Solution

(a) becuase there is a steady speed downwards, do i just use gravity as the accelleration and nothing else? so F= (80) x (10) F=800 N

/or/ is it F= (80) x (10+2) F=960 N

The first.

One thing that is wrong with this second possibility is that you are trying to add an acceleration (10 ms^-2) to a velocity (2 ms^-1). This is never possible - in general you can't add or subtract quantities with different dimensions.

(b) F= (80) x (10-4) so F= 480 N

This says that the object appears lighter when the downward-moving lift comes to a halt. Does that tally with your everyday experience?

The way to solve this sort of problem is not to randomly plug numbers into an equation in the hope that it might be vaguely suitable. First draw a free-body diagram showing all the forces on the body you're considering. Then you can use Newton's second law, like you suggest: the sum of the forces is the mass times the acceleration - but this isn't very helpful unless you have carefully considered all the forces!

(c) F= (80) x (10) F=800

Correct - can you explain why? What is the crucial feature this case has in common with (a)?

 

[druuuuuuuunnk]

THANKS FOR YOUR HELP FIRST OF ALL...

(c) & (a) both have only gravity acting on them as there acceleration?

(b) good point about the overall weight being lighter, i should have thought about it when i saw it was 480N.

So when the lift is coming to a stop you add to the overall acceleration and when the lift is accelerating downwards you take away, but if it was accelerating/decelerating upwards you would also add? becuase either way the force is pushing on the object.

thanks also for tell me about how you can't add things of a different dimension, its simple rules like that which my teacher has neglected to teach. you've saved me a lot of time and trouble!