Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating normal force and net force of 20 kg mass going down inclined plane at

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    angles of 30, 45, 60. calculate weight of block, the normal force, and net force for each angle separately.

    2. Relevant equations
    normal force m*g*sin theta
    net force m*g*cos theta

    3. The attempt at a solution
    weight of block is just 20 kg, right?

    and so for 30 degrees: normal force= 20 kg* 9.81 m/s/s * sin 30, right? and net force=20 kg * 9.81 m/s/s * cos 30...
  2. jcsd
  3. Oct 24, 2007 #2
    No. The weight of an object is measured in lbs or N. kg is a unit of mass.

    Weight in Newtons = mass in kg*acceleration due to gravity in m/s^2.

    Draw a diagram showing all your forces.
  4. Oct 24, 2007 #3
    wait....my next question then asks how does each force value change as the angle changes and why?

    in this case, wouldn't the weight be constant...?
  5. Oct 24, 2007 #4
    You're absolutely right. (weight is constant). Your calculations are not, however, correct.
  6. Oct 24, 2007 #5
    ok, then can you please tell me what i did wrong and how to calculate them correctly?
  7. Oct 24, 2007 #6
  8. Oct 24, 2007 #7
    ugh. i'm still confused.
  9. Oct 24, 2007 #8
    Gm is the weight, N is the normal force. the Alpha sign is the angle. The same angle of the incline. Do you see that a triangle has formed? You can apply the trig functions of sine and cosine.
  10. Oct 24, 2007 #9
    so normal force= 196.2 N * cosine 30=170N, which is what I had,
    and net force= 196.2 N *sin 30=98.1

  11. Oct 24, 2007 #10
    is that correct?
  12. Oct 24, 2007 #11
    You had normal force = 196.2sin30 which is not correct.

    Now your calculations look correct. If there was friction then the net force would be the force going down the incline subtracted from the friction force(going up the incline).

    But everything looks good! great job!
  13. Oct 24, 2007 #12
    right, i did have it like that and just now after i looked at my notes, i realized i accidently switched the two formulas.

    thanks for your help. i really appreciate it!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook