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Find the net force with the given position vector

by dimeloom
Tags: net force, position vector
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dimeloom
#1
Sep13-10, 01:04 PM
P: 2
[b]1.The position vector of a 3000N helicopter is given by:
r=(.02 m/(s^3))(t^3)i + (2.2m/s)tj - (0.06 m/(s^2))(t^2)k
Find the net force on the helicopter at t = 5s.



[b]2. I really have no clue where to begin.



[b]3. I started doing this:
r= (.02m/(s^3))(5s^3)i + (2.2m/s)(5s)j - (.06m/(s^2)*(5s)^2k
r= 2.5mi + 11m + 1.5m
then I drew a 3d graph with each axis labeled and two right triangles but have no clue
where to go from here.
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rl.bhat
#2
Sep14-10, 07:12 AM
HW Helper
P: 4,433
The given problem can be written as

r = a*t^3*i + b*t*j -c*t^2*k

Find d(r/dt and d^2(r)/dt^2. That gives you the acceleration. Find the magnitude of a.

Weight of helicopter is given. Find its mass. Then the force on the helicopter = m*a
dimeloom
#3
Sep14-10, 02:18 PM
P: 2
Thank you so mu rl.bhat.

After the 2nd derivative I get a= (.12m/s^3)(t) - .12m/s^2
= (.6m/s^2)- .12m/s^2= -.6 m/s^2
F=m*a m=F/a m=3000n/9.81m/s^2=305.8kg

m*a=F 305.8 kg * -.6m/s^2=-183.48N Is that thought process correct?

rl.bhat
#4
Sep14-10, 08:41 PM
HW Helper
P: 4,433
Find the net force with the given position vector

After the 2nd derivative I get a= (.12m/s^3)(t) - .12m/s^2

It should be

a= (.12m/s^3)(t)*i - .12m/s^2*k

Find the resultant a.


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