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Two people pulling a car please help.

  1. Sep 1, 2007 #1
    Two forces are applied to a car in an effort to move it, as shown in Figure P4.12.
    Here's Figure P4.12: <img src="http://www.webassign.net/sercp/p4-12.gif"/>

    (a) What is the resultant of these two forces?

    Magnitude: N
    Direction: Degrees (measured to the right from the forward direction)

    (b) If the car has a mass of 3000 kg, what magnitude acceleration does it have? Ignore friction.


    For part A, I found F1 = (400cos30)i + (400sin30)j
    F2 = (450cos10)i + (450sin10)j

    ___+______________________________
    Fnet=[346i + 443i] + [200j + 78.1j]
    Fnet= 789i + 278.1j

    I then used the Pythagoreon Thereom to figure out the resultant force. I got 837 N. However, webassign.net says this is wrong and I've been re-doing the problem over and over and get the same answer.

    I used tan-1 [278/789] for the angle and got 19.4 degrees but webassign.net also tells me this is the wrong answer.

    Finally, for part B, I used F=ma and used 837=3000a and solved for a. I got a=.279 m/s^2 but again webassign.net tells me this wrong.

    So, basically, I was hoping someone could look through my work and tell me if you can figure out what I'm doing wrong cause I could swear I did this correctly.
     
  2. jcsd
  3. Sep 1, 2007 #2

    learningphysics

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    Your force equations aren't right... take i in the positive x direction, and j in the positive y direction...

    write out the equations again... note that one force has a positive x component, the other force has a negative x component...
     
  4. Sep 1, 2007 #3

    Doc Al

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    Staff: Mentor

    Define the i & j directions. Are they horizontal and vertical? Or the reverse?

    Be careful with signs.
     
  5. Sep 1, 2007 #4
    So, instead of using 450 I should be using -450?
     
  6. Sep 1, 2007 #5

    learningphysics

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    sketch the forces and their components... what is the direction of the x-component of the 450N force... what is direction of the y-component of the 450N force
     
  7. Sep 1, 2007 #6
    Ok, so now I've changed my Equations so they are:

    Fnet=[346i+ -443i] + [200j + 78.1j]

    For a total of -97.0i + 278j

    Am I doing any better at this or I'm still doing it wrong??
     
  8. Sep 1, 2007 #7

    learningphysics

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    No, it's still wrong... can you define your directions? I'm confused by what directions i and j are...
     
  9. Sep 1, 2007 #8
    i is in the horizontal (x) direction and j is in the vertical (y) direction.

    At this point I'm about to just scrap all my work and start over. I'm thinking that might be easier than going back and making corrections.
     
  10. Sep 1, 2007 #9

    learningphysics

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    I agree, start fresh... and be careful with directions...
     
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