# vertical motion mechanics problem

• Mar 24th 2010, 01:07 AM
furor celtica
vertical motion mechanics problem
a lift starting from rest moves downward with constant acceleration. It covers a distance s in time t, where s = (1/6)(g)(t^2). A box of mass m is on the floor of the lift. find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.
i got 5/6(mg) and 11/12(mg) using a= d/t^2 and v^2 = u^2 + 2as but the answer is supposed to be 2/3(mg)
what is wrong with these formulae? i dont think i made any calculation mistakes
• Mar 24th 2010, 05:12 AM
sa-ri-ga-ma
Using the kinematic equation
s = vo*t + 1/2*g*t^2
you can see that the acceleration of the lift is 1/3*g.
The apparent acceleration of the mass inside the falling lift is
g' = (g -1/3*g) = 2g/3.
So the reaction of the floor on the mass is 2mg/3
• Mar 24th 2010, 05:30 AM
furor celtica
how were my formulae inappropriate?
• Mar 24th 2010, 06:10 AM
sa-ri-ga-ma
Quote:

Originally Posted by furor celtica
how were my formulae inappropriate?

How did you get 5/6 mg and 11/12mg? Show your calculations.
From where did you get a = d/t^2?
• Mar 24th 2010, 10:34 PM
furor celtica
good point, where the hell did i get that?
i guess i just developed from the general form of acceleration values, you know distance/time^2 but i guess thats wrong huh
how exactly?
• Mar 25th 2010, 12:15 PM
running-gag
$\vec{P} = m\vec{g}$ and $\vec{N}$ are the 2 forces applying on the box

$m\vec{a} = m\vec{g}+\vec{N}$

Using an y-axis oriented downwards with y=0 at time t=0

$ma_y = mg-N$

The acceleration of the box is the same as the lift one $\frac{d^2s}{dt^2} = \frac{1}{3}g$

$\frac{1}{3}mg = mg-N$

$N = \frac{2}{3}mg$