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Homework Help: Forces and Vectors Problem?

  1. Sep 12, 2004 #1
    First here is the problem: Only two forces act on an object(mass 3.00kg) as in the drawing(40.0N in the x direction and 60.0N north-east and a 45 degree angle between them) Find the magnitude and direction, relative to the x-axis of the acceleration of the object.

    I have tried finding the x and y components of each vector and then adding the x components and y components together to get the resultant and then using a=F/M and the tangent to determine the angle(direction) BUT I keep getting the wrong answer!! Please explain!!
     
    Last edited: Sep 12, 2004
  2. jcsd
  3. Sep 12, 2004 #2
    Did you get this for the components : first force = 40e_x
    second force = 60(cos(45)e_x + sin(45)e_y) and cos(45) = sin (45)= 1/sqrt(2)???

    e_x and e_y just denote the basis-vectors...

    If you did get this, you are right...

    Remeber that is this exercise you only need to add the x-components in order to find the answer. We are only looking in the x-direction here..
     
  4. Sep 12, 2004 #3
    Shouldn't the answer be 27,47546896 m/s² ???
     
  5. Sep 12, 2004 #4
    That is how I did the problem, but the answer in the book is a=30.9(m/s2) and the degree is 27.2 degrees above the x-axis??!!
     
  6. Sep 12, 2004 #5
    Ahh, got it

    They are asking the resulting force when you add the two given forces together

    use the cosinus-rule...
     
  7. Sep 12, 2004 #6
    F² = 40² + 60² -2*40*60*cos(135) Then take the squareroot and devide by three and you will get 30,9 m/s²
     
  8. Sep 12, 2004 #7
    Or with the components

    (total force)² = (60/sqrt(2) + 40)² + (60/sqrt(2))² = 92,7

    The angle is calculated using tan(x)=Y/X and Y component = 60/sqrt(2)
    and X component = (40 + 60/sqrt(2))

    regards
    marlon
     
  9. Sep 12, 2004 #8

    Pyrrhus

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    Homework Helper

    Do what marlon says or

    [tex] R_{x} = 60cos(45) + 40cos(0) [/tex]

    [tex] R_{y} = 60sin(45) + 40sin(0) [/tex]

    [tex] \vec{F_{R}} = (82.42i + 42.42j) N [/tex]

    [tex] \vec{A} = (\frac{82.42}{3}i + \frac{42.42}{3}j) m/s^2 [/tex]

    [tex] \vec{A} = (27.473i + \14.14j) m/s^2 [/tex]

    [tex] |\vec{A}| = \sqrt{27.473^2+14.14^2} [/tex]

    [tex] |\vec{A}| = 30.89 [/tex]

    [tex] \theta_{A} = arctan(\frac{14.14}{27.473}) [/tex]

    [tex] \theta_{A} = 27.23^o [/tex]
     
  10. Sep 12, 2004 #9
    Thanks! I understand I was using cos(45) not cos(0)!
     
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