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: Please help with calculus problem

  1. Apr 11, 2008 #1
    y axis = force
    x axis = distance
    xy = torque
    I have two points on an x axis
    0 and 6
    at 0 xy = -18.352
    at 6 xy = .802
    I need to know what y equals and its value of x. How can I do this?
     
  2. jcsd
  3. Apr 11, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I have no idea what you are talking about. At distance=0 torque=-18.352? What torque? What distance?
     
  4. Apr 12, 2008 #3
    nevermind
    i solved it

    (6-z) x F = .802
    z x F = -18.352
     
  5. Apr 12, 2008 #4
    can the above 2 equations be solved?
     
  6. Apr 12, 2008 #5

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    If x is a variable, you have two equations in three unknowns which cannot be solved uniquely. If x denotes multiplication, you have two equations in two unknowns so there is usually a unique solution. If it is an outer product ([itex]\times[/itex]) the equations make no sense.
     
  7. Apr 12, 2008 #6
    what I want to know is what force magnitude and what point along the x axis does the force have to act on to make the fulcrum at x = 0 and the fulcrum at x = 6 have the following torques
    at 0 = -18.352
    at 6 = .802
     
  8. Apr 12, 2008 #7
    what do you mean by outer product ([itex]\times[/itex]) ?
     
  9. Apr 13, 2008 #8
    Statement of problem

    I assume that all distances are measured on the x axis from x=0, negative to the left and positive to the right. Also I assume that force is measured positive downward.

    The downward force generating a torque of -18.352 about 0 must be applied to the left of 0.
    The downward force generating a torque of 0.082 about 6 must be applied to the right of 6.
    Therefor the problen has no solution or I have stated it incorrectly.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook