How Do You Solve This Calculus Problem Involving Torque and Force?

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Homework Help Overview

The discussion revolves around a calculus problem involving torque and force, specifically how to determine the force magnitude and its application point along the x-axis to achieve specified torques at two points. The original poster presents two torque values at distances 0 and 6 on the x-axis.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between torque, force, and distance, questioning the definitions and assumptions regarding the variables involved. There are attempts to clarify the nature of the equations presented and whether they can be solved uniquely.

Discussion Status

The discussion includes various interpretations of the problem setup and the equations provided. Some participants express confusion about the terms used, while others attempt to clarify the conditions under which the equations can be solved. There is no explicit consensus on the problem's formulation or solution.

Contextual Notes

Participants note assumptions about the measurement of distances and forces, including the direction of the forces and the positioning of the torques. There are indications that the problem may be incorrectly stated or lacks sufficient information for a definitive solution.

Ry122
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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?
 
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I have no idea what you are talking about. At distance=0 torque=-18.352? What torque? What distance?
 
nevermind
i solved it

(6-z) x F = .802
z x F = -18.352
 
can the above 2 equations be solved?
 
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 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
 
CompuChip said:
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]) ?
 
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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