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Simple Statics problem I am having brain fart on

  1. Sep 29, 2011 #1
    I know I am missing something simple here. This is only the second problem and I have done the rest with no problems. I would appreciate a second set of eyes.
    1. The problem statement, all variables and given/known data

    The mass center G of the 1430-kg rear-engine car is located as shown in the figure. Determine the normal force under the left front tire (a)Nf (point F) and under the left rear tire (b)Nr (point R) when the car is in equilibrium.

    F is the front tire. Center of gravity is 1406mm to the right of F. R is rear tire which is 972mm from CG or 2378 to the right of F.

    2. Relevant equations

    I know you can use 3 different equations to solve these 2D static problems. Sum of forces in x direction, sum of forces in y direction, and sum of the moment about a point of your choosing.

    3. The attempt at a solution
    All the forces are normal to the x-axis so there are no forces acting in the x direction. Summing in the y-direction gives you two unknowns of F(front) and R(rear).
    I chose to sum the moment about F. The normal force at F slides through the point so the it's contribution is zero.

    M about F= (9.81m/s^2)(1430kg)(1.406m)-R(2.378m)=0
    I chose clockwise to be positive.
    The answer I get for R is 8294N but apparently is wrong. I am guessing I am over looking something simple.
    Thanks for the help!
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2

    PhanthomJay

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    First clarify the CG....is it 1423 or 1406 mmm to the right of F?

    Your equation otherwise would appear correct, except that most cars have 4 tires, not 2. :wink:
     
  4. Sep 29, 2011 #3
    woops sorry. The CG is 1406mm to the right of the front tire.
    Interesting side note. I found an similar example with an answer and was able to find out that my answer was double the correct answer. Now I am even more confused:)
     
  5. Sep 29, 2011 #4

    PhanthomJay

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    your answer assumed that the car had 2 tires..one front and one back....but the car has 4 tires, 2 front and 2 back, so the normal force you calculated is shared by 2 tires...that's why your answer is off by a factor of 2.......
     
  6. Sep 29, 2011 #5
    Wow thanks. I knew it was a brilliant mistake somewhere.
     
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