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Please help with calculus problem

  • Thread starter Ry122
  • Start date
565
2
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?
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
I have no idea what you are talking about. At distance=0 torque=-18.352? What torque? What distance?
 
565
2
nevermind
i solved it

(6-z) x F = .802
z x F = -18.352
 
565
2
can the above 2 equations be solved?
 
CompuChip
Science Advisor
Homework Helper
4,284
47
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.
 
565
2
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
 
565
2
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.
what do you mean by outer product ([itex]\times[/itex]) ?
 
70
0
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.
 

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