apparent weight

[thatricksta]

ok so, ive been doing a heap of exams and have found some sort of an inconsistency with the calculation of apparent weight. (normal reaction force)

Rather than giving me the equation, id like to know if there would be a difference between calculating the normal force at the top of a circle for a plane (doing a vertical circle) or for a car going over a speed hump.

my understanding is N = mg - Fc

thanks

[HCbigstick]

On a speed hump, N = mg - Fc whereas at the top of a rollercoaster (inside), N = Fc - mg

[Comma]

Yes there is a difference.

For a car going over a speed bump, the net force is downwards and the normal reaction force is upwards
For the top of a circle for a plane, the net force is downwards, as is the normal reaction force. This then gives you N = Fc - mg

[adelaide.emily10]

for a lift, when going up you are suppose to feel heavier then how come in vcaa 09 the normal reaction force is less than a weight force (i understamd the working out) but just slightly confused in feeling lighter or heavier.

[HCbigstick]

for a lift, when going up you are suppose to feel heavier then how come in vcaa 09 the normal reaction force is less than a weight force (i understamd the working out) but just slightly confused in feeling lighter or heavier.

Depends whether or not you are accelerating/decelerating.

[adelaide.emily10]

so if you start from rest and accelerate up you will find that the apparent weight is equal to the weight force plus the force provided by the acceleration (f=ma) so you feel heavier
but if you are at a constant spped and then decelerate you will feel lighter because acceleration is negative?

[HCbigstick]

Correct.