# Find the net force with the given position vector

by dimeloom
Tags: net force, position vector
 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.
 HW Helper P: 4,430 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
 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?
HW Helper
P: 4,430

## 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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