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  1. Nov 16, 2003 #1
    4) A rescue helicopter is lifting a man (weight- 822 N) from a capsized boat by the means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.10m/s^2? (b) What is the tension during the remainder of the rescue when he is pulled upward at constant velocity?

    a) The mass of the man is 83.9kg

    T = (83.9Kg)(9.80m/s^2+1.10m/s^2)

    T = 914.5N

    b) It is 822N because the velocity is constant.

    5) Three forces act on a moving object. One force has a magnitude of 80.0N and is directed due norht. Another has a magnitude of 60.0N and is directed due west. What must be the magnitude and direction of the third force, such as the object continues to move with constant velocity.

    I know how to find the magnitude, but have trouble finding the direction. Someone suggested tan(angle)= x/y. But when I do it out out i get a small number that couldn't possibly be the angle.

  2. jcsd
  3. Nov 16, 2003 #2

    Doc Al

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    Staff: Mentor

    You did problem 4 OK.
    Show what you did. (You must have made an error somewhere!)
  4. Nov 16, 2003 #3
    Hi, thanks for your help.

    I think I realized what I did wrong. You have to turn the formula around to solve for just the angle so its

    angle = (80N/60N)/tan = 57.2

    Do I answer the direction is 57.2 degrees North West?
  5. Nov 16, 2003 #4
    No, the third force is opposite to the sum of the first two (constant velocity means no net force).

    Also, 57.2 degrees is not the arctangent of 4/3 (=80/60). I don't know what you did to get that angle.

    By the way, you have to be careful to specify the reference angle. Is it 57.2 degrees from the north axis, or the west axis, or what?
  6. Nov 16, 2003 #5
    angle = (80N/60N)/tan = 57.2

    Thats how I got the previous answer. How do I calculate it the right way?

    And I guess it would be from the X axis since the y-component of the force is greater.
    Last edited: Nov 16, 2003
  7. Nov 16, 2003 #6
    What does that even mean? tan of what? tan isn't a variable, it's a function. And why are you dividing by it? That doesn't make mathematical sense.

    tan(angle) = (y/x)
    angle = arctan(tan(angle)) = arctan(y/x)

    Yes, it's measured clockwise from the negative-x (west) axis.
  8. Nov 16, 2003 #7
    Last question..

    So what do I do with the 4/3 take the "arctan" of it?

  9. Nov 16, 2003 #8
    You use the arctangent function on your calculator.
  10. Nov 16, 2003 #9
    53.1 degrees from west axis. I hope this sounds right. Thanks again for all your help!
  11. Nov 16, 2003 #10
    Yes, that's for the sum of the first two forces. Remember that the third force is opposite to that (53.1 degrees south of the east axis).
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