- #1

Runaway

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## Homework Statement

A machine in an ice factory is capable of exerting 3.3 * 10^2 N of force to pull large blocks of ice up a slope. The blocks each weight 1.21 *10^4 N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is to be able to complete the task.

Fg= 1.21*10^4 N

Ft= 3.3 * 10^2 N

Fn is perpendicular to Ft

## Homework Equations

(in case the symbols my teacher uses are not standard)

Fg= weight

Ft= tension force

Fn= normal force

F()x= the x component of said force ie Ftx

F()y= the y component of said force ie Fty

## The Attempt at a Solution

These are the only equations I can come up with using the given data, but no matter how I substitute them, I have yet to find a solvable equation.

Ftx= Ft/sin([tex]\theta[/tex])

Fty=Ft/cos([tex]\theta[/tex])

Fnx=Fn/sin(90-[tex]\theta[/tex])

Fny=Fn/cos(90-[tex]\theta[/tex])

Ft= sqrt(Ftx

^{2}+ Fty

^{2})

Fn= sqrt(Fnx

^{2}+ Fny

^{2})

Fty+Fny=Fg

Ftx=Fnx

but no matter how I substitute them, I have yet to find a solvable equation.

ex.

1.21*10^4 N=(3.3 * 10^2 N)/cos([tex]\theta[/tex])+(sqrt(((3.3 * 10^2 N)/sin([tex]\theta[/tex])

^{2}+ ((1.21*10^4 N)-(3.3 * 10^2 N)/cos([tex]\theta[/tex]))

^{2}))/cos(90-[tex]\theta[/tex])

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