What is the Net Force on a Helicopter at 5 Seconds?

[xmflea]

The position of a 2.75x10^5N helicopter under test is given by

r=(0.020m/s^2)t^3i + (2.2m/s)tj - (0.060m/s^2)t^2k

Find the net force of the helicopter at t=5.0s express the vector F in the form Fx, Fy, and Fz

so far i plugged in 5.0 to each component in the r equation which gave me different masses: 2.5, 11.0, and 1.5

then i divided each mass by 5.0 to get velocity, then i plugged everything into the average accleration formula to get different accelerations, and then i finally used F=ma to plug into get different forces for Fx Fy and Fz
.. and it looks like I am just going in a huge circle. not sure how to do this.

 

[Kurdt]

How do you find acceleration when given position as a function of time?

 

[baba89]

I would approach this problem by first understanding the basic principles of Newton's laws of motion. In this case, since we are dealing with a helicopter, we would need to consider the forces acting on it, such as gravity, air resistance, and the force generated by the helicopter's engines.

To find the net force of the helicopter at t=5.0s, we can use the formula F=ma, where F is the net force, m is the mass, and a is the acceleration. However, in order to use this formula, we need to first calculate the acceleration of the helicopter at t=5.0s.

To do this, we can use the given position equation r=(0.020m/s^2)t^3i + (2.2m/s)tj - (0.060m/s^2)t^2k and take the second derivative with respect to time to find the acceleration. This would give us a=0.120i + 2.2j - 0.120k m/s^2.

Now, we can plug in this acceleration value into the formula F=ma to find the net force. Since the mass of the helicopter is not given, we can assume it to be 2.75x10^5N (given in the problem statement). This would give us a net force of F=(0.330i + 605j - 0.330k) x 10^5 N at t=5.0s.

In order to express this vector in the form Fx, Fy, and Fz, we can simply break down the vector into its x, y, and z components. This would give us Fx=0.330 x 10^5 N, Fy=605 x 10^5 N, and Fz=-0.330 x 10^5 N.

In conclusion, the net force of the helicopter at t=5.0s is (0.330i + 605j - 0.330k) x 10^5 N or (0.330 x 10^5 N, 605 x 10^5 N, -0.330 x 10^5 N) when expressed in its x, y, and z components.