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How do you find frictional force with no mass given?

  1. Dec 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Batman is driving the Batmobile down a hill coming from the Bat Cave. This hill is inclined at an angle of 30 degrees to the horizontal and has a coefficient of kinetic friction of 0.28. What force must the Batmobile's engine apply to cause the Batmobile to accelerate at 0.60g?

    a = 0.60g = 0.60(9.8) = 5.88
    uk = 0.28


    2. Relevant equations
    Fengine = ma + mg sin30 + Ff


    3. The attempt at a solution
    I tried plugging in what I had into the above equation, assuming that the mass was 0.60, so I got Fengine = (0.6)(5.88) + (0.6)(9.8)sin30 + 1.64 and my answer was 8.1 N as the force applied on the engine. But when I looked at the back of the book, it said that the answer is (3.36)(mass) N. How does that work? I am so confused!
     
  2. jcsd
  3. Dec 22, 2009 #2

    ideasrule

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    Homework Helper

    0.6 kg is about the weight of five apples, whereas the Batmobile is the size of a tank. How can you assume that m is 0.6 kg? Just call it "m" and assume it's known.

    First, draw a free-body diagram of the Batmobile. Remember to label all forces. Then write Newton's second law for the direction parallel to the ramp and the direction perpendicular to it.
     
  4. Dec 24, 2009 #3
    I'm also quite interested in this question, i come up with two unknowns and only one equation.
     
  5. Dec 24, 2009 #4
    I attempted this question, i myself am reviewing for my standardized year-end tests..

    though im not sure it is correct.

    i stated that:

    [tex]F_{engine} = F_{net} + F_F - F_x[/tex]

    Where Fx and Fy are the components of the gravitational force on the object.

    After substitution:

    [tex] F_{engine} = ma + \mu mg \cdot cos\Theta - mg \cdot sin\Theta[/tex]

    Then i took the mass as the common factor:

    [tex] F_{engine} = m\left( a + \mu g \cdot cos\Theta - g \cdot sin\Theta\right)[/tex]

    And then i solve:
    [tex] F_{engine} = m\left( (0.60)(9.8) + (0.28)(9.8)(cos(30)) - (9.8)(sin(30))\right) [/tex]

    i end up with [tex]F_{engine} = 3.37 \cdot m[/tex] Newtons

    which is as much as i could simplify, i dont think you can fully solve this question without the mass.. am i correct?
     
  6. Dec 24, 2009 #5
    yes, unless the acceleration could be provided by gravity itself:

    i.e sin(a) - u cos(a) = 0.6 Your answer is in agreemnet with that in the book according to the OP.
     
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