Posted by QUESTION on Thursday, August 20, 2009 at 7:45am.
"acceleration first half of trip
i got 5.880 s^-2 m "
OK, as it equals 0.6g, the maximum acceleration without the box slipping from the belt.
I do not know how you got the time for the first half, my formula is
S = ut + (1/2)at²
for initial velocity u=0, a=5.88, and distance S=10/2=5m
t=√(2S/a)=10/5.886=1.303 s.
So the minimum time is twice this value, 2*1.303=2.606 s., because the deceleration takes the same time as the acceleration.
The boxes may slip a little forward as the acceleration changes to deceleration, but that will only make the box arrive a fraction of a second earlier.
Acceleration for first half of trip...
apply newtons second law...
net force = m a = - F fr = - (Mu s Fn)
were Mu s is coefficent of static friction
take the inverse of m onto both sides
a = m^-1 -(Mu s Fn)
apply Newtons second law in the y direction
net force (y direction) = ma = FN - Fg = 0
0 sense it's not moving in this direction and FN is the normal force
add Fg to both sides
FN = Fg = mg
plug this value of FN into the equation for Newtons Second Law in x direction which was rearanged for a...
a = m^-1 -(Mu s Fn)=m^-1 -(Mu s mg)
cross out the masses
a = - (Mu s g )= -(.60 (-9.80 s^-2 m))
two negatives cross out giving positive
a = 5.880 s^-2 m
X = Xo + Vo t + 2^-1 a t^2
Xo is zero and so is Vo
X = 2^-1 a t^2
solve for t
t = (a^-1 2 x)^(2^-1)
t = (5.880 s^-2 m (2) 50 m)^(2^-1)
t = 4.124 s
velocity as half
v = (2a (X - Xo) + Vo^2)^(2^-1)
Xo is zero and so is Vo^2
v = (2a X)^(2^-1)
v = (2 (5.880 s^-2 m) 5.0m)^(2^-1)
v = 7.668 s^-1 m
acceleration second half
a x = (V^2 - Vo^2) + Xo
solved for a were V^2 is zero
a = x^-1(-Vo^2 + Xo)
a = (10 m)^-1 (-(7.668 s^-1 m)^2 + 5.0 m)
a = -6.380 s^-2 m
time for second half
a = t^-1 (V - Vo)
t = a^-1 - Vo
t = (- 6.380 s^-1 m)^-1 -(7.668 s^-1 m)
t = 1.202 s
total t = 1.202 s + 4.124 s
t = 5.3 s
how do you incorporate that theres is static friction i thought I did
The static friction is incorporated, as mathmate indicated, by assumeing the max acceleration and deaccelearation was .6 g.
Fmax=fn*mu= mg*.6
Fmax=m(accelerationmax)=mg*.6
accelerationmax=.6g
I did that
if you do this
t=ã(2S/a)=10/5.886=1.303 s.
your doing the full distance 10 m instead of 5 m right???
so why would you double it
I believe there is a discrepancy of the data.
According to the posted question,
"Boxes are moved on a converyor blet from where they are filled to the packing station 10m away..."
with which 1.304 seconds for half of the trip was obtained, using the formula
t=sqrt(2*S/a) where S=half of total distance and a=acceleration.
Your calculation probably assumes a half-distance of 50m to get
"t = (5.880 s^-2 m (2) 50 m)^(2^-1)
t = 4.124 s "
The second half of the trip should take exactly the same time as the first half, whether the distance is 100 or 10 m.
the doubling comes from:
S=ut+(1/2)at²
when u=0,
S=(1/2)at²
t²=2*s/a
t=sqrt(2*s/a)
oh wow...
Thanks!!!
You're welcome!