Vectors and Angles in Helicopter Flight

  • Thread starter Thread starter gric122
  • Start date Start date
  • Tags Tags
    Angles Vectors
AI Thread Summary
Bob's helicopter flight involves three segments with specific headings and distances, leading to a complex displacement calculation. The first leg is 17 kilometers at 63 degrees East of South, followed by 45 kilometers at 73 degrees West of North, and finally 34 kilometers at 79 degrees West of North. The main challenge lies in accurately calculating the resultant displacement vector by breaking down each leg into north/south and east/west components. There is confusion regarding the calculation of the x and y components, particularly in using trigonometric functions correctly for angles in degrees. The discussion emphasizes the importance of correctly applying sine and cosine functions to determine the final displacement needed to return to the launch pad.
gric122

Homework Statement



Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 17 kilometers at heading of 63 degrees East of South. Then he flies 45 kilometers heading 73 degrees West of North. After this he flies 34 kilometers heading 79 degrees West of North. Now he is ready to return to the launch pad.

What is the displacement vector that he needs to take to return directly to the launch pad from his present location (for the heading give the number degrees north of east - your answer may be greater than 90 degrees)?

Homework Equations


none

The Attempt at a Solution


So I've attempted to draw this problem out and then solve it with trig, but I'm stuck on where the angles point to. From what I've drawn, it looks like a slanted parallelogram, and I don't think I can trig it
 
Physics news on Phys.org
I would just work out the displacement in the north/south and east/west direction for all the given distances, and then add those all up.
The final leg of the jouney will be in the opposite direction of the result.
 
I calculated the horizontal distance and got -61.26 km, but for some reason my vertical is 8.66 which seems really low, am I doing somethign wrong? For x I did 17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
 
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
When I do that I get 11.9.
 
haruspex said:
When I do that I get 11.9.
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
 
gric122 said:
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
I'm referring to the y value, obtained as you posted:
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
I.e. just replace all of those cos functions with sine functions. And, yes, angles in degrees.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top