Apparent Weight (1 Viewer)

[elseany]

when i was writing up my notes, i ran into quite a stump in trying to explain what apparent weight was. Like i know that mathematically it is ma + mg and i get the mg part being the weight factor but i dont get the ma part.

I think i kind of understand it and what im thinking is that its the sum of the true weight and the reaction force. Is this correct?

The thing that is confusing me is this sheet we got handed out in class, i dont have a scanner so i quickly redid the diagram in paint:

and next to it this is the working out they give (letters in brackets are subscript):

Net force = centripcial force = F(n) = ma(c) = mv^2/r = R + W = R + mg

R = mv^2/r - mg

g-force = R/W = (mv^2/r - g)/mg = (v^2/r - g)/g

okay whats going on? It was my understanding from the diagram that R is equal to the reaction force but now there saying that the reaction force is the apparent weight (this is the part that confuses me since i thought the apparent weight is the sum of the true weight and the reaction force [or the difference in this case]).

if the above is illegible blabber, please disregard it and explain what apparent weight means please :<

 

[Wackedupwacko]

not necessarily... apparant weight... if you want to define it in simple terms its what you would feel if you were in that situation. like if your in an elevator and u accelerate up you feel slightly *heavier* (if it accelerates fast enoguh and long enough).

so yes at times its mg + ma for the apparant weight. the ma is the acceleration up (it must oppose the direction of gravity) . that elevator accelerating up is one of those examples, rocket accelerating into space is another.

the actual formula i believe for apparnt weight is ma-mg (for g force just divide by mg) and thus when u freefall you feel weightless since a = g .

however in the case of circular motion you have the centrifugal force spinning out opposite R(and W) however they are still moving in that circlur motion so the forces opposing it must be the same in magnitude but opposite in direction.

SO: mv^2/r - (R+W) = 0
R+W = mv^2/r
R = mv^2/r -W
so R = m(v^2/r -g)

and thus R the aparant weight is mv^2/r - mg which still is true since a = v^2/r in circular motion.

hope that cleared things up