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Question about friction and normal force

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A 3.5 kg block is pushed along a horizontal floor by a force F of magnitude 15 N at an angle of 40 degrees DOWNWARD. The coefficient of kinetic friction between the block and the floor is 0.25.

    My question is about the normal force. I know the normal force is given by:

    N - mg - Fsin(theta) = 0.

    N = mg + Fsin(theta)

    The correct normal force they obtain in the problem is from (3.5)(9.8) + 15sin(40)

    I am wondering why they chose the angle as 40 when it was directed down into the 4th quadrant. Why isn't the angle then -40 degrees (negative 40 degrees) because it is in the 4th quadrant, below the horizontal.

    Picture: http://www.google.com/imgres?q=a+3....=rc&dur=655&sig=111188412953169759186&page=1&

    2. Relevant equations

    N = mg + Fsin(theta)

    3. The attempt at a solution

    (3.5)(9.8) + 15sin(40)
    (3.5)(9.8) + 15sin(-40)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 4, 2012 #2
    When you choose "-" sign for the third term (first equation of your post) you already took into account the fact that the force is downwards, the angle is in the fourth quadrant etc.
  4. Feb 4, 2012 #3
    Thanks nasu. Much appreciated
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