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"One end of a uniform meter stick is placed against a vertical wall View Figure . The other end is held by a lightweight cord that makes an angle [itex]\theta[/itex] with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.400 ."
I can't for the life of me get the equation for this.
Part A:
What is the maximum value the angle [itex]\theta[/itex] can have if the stick is to remain in equilibrium?
So far I've tried using the last end as a pivot point.
so that would give me
Balancing torque:
[tex]\mu T cos(\theta) = mg(1-x)[/tex]
and balancing total forces
[tex]\mu T cos(\theta) + T sin(\theta) = mg [/tex]
combining the two
produces
[tex] T sin(\theta) = mgx[/tex]
dividing the third by the first gives me
[tex] tan(\theta) = \mu \frac{x}{1-x} [/tex]
and there is no way I can get a "max theta". Naturally I can't get Part B and C but I figured if you guys can help me with A it will be easy to get the rest.
Thanks in advanced.
"One end of a uniform meter stick is placed against a vertical wall View Figure . The other end is held by a lightweight cord that makes an angle [itex]\theta[/itex] with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.400 ."
I can't for the life of me get the equation for this.
Part A:
What is the maximum value the angle [itex]\theta[/itex] can have if the stick is to remain in equilibrium?
So far I've tried using the last end as a pivot point.
so that would give me
Balancing torque:
[tex]\mu T cos(\theta) = mg(1-x)[/tex]
and balancing total forces
[tex]\mu T cos(\theta) + T sin(\theta) = mg [/tex]
combining the two
produces
[tex] T sin(\theta) = mgx[/tex]
dividing the third by the first gives me
[tex] tan(\theta) = \mu \frac{x}{1-x} [/tex]
and there is no way I can get a "max theta". Naturally I can't get Part B and C but I figured if you guys can help me with A it will be easy to get the rest.
Thanks in advanced.
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