Need help on static friction problem

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SUMMARY

The discussion centers on calculating the coefficient of static friction for a framed picture weighing 5.0 N pressed against a wall with a force of 6.0 N at a 40º angle from the vertical. The equations used include ƩFx=0 and ƩFy=0, leading to the formula μ=(6cos40°-5)/6sin40°. The analysis reveals two possible directions for the frictional force: upward when the picture tends to slide downward and downward when the picture tends to slide upward. The problem is resolved by understanding the conditions under which static friction acts in both scenarios.

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yunny
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Homework Statement



The problem is
While trying to decide where to hang a framed picture, you press it against the wall to
keep it from falling. The picture weighs 5.0 N and you press against the frame with a
force of 6.0 N at an angle of 40ºfrom the vertical.
What is the coefficient of static friction between the wall and the picture? The
frictional force exerted on the picture by the wall can have two possible directions.
Explain why.


Homework Equations



ƩFx=0
ƩFy=0
f≤μN

The Attempt at a Solution



I used ƩFx=0,
f=6cos40°-5
ƩFy=0,
N=6sin40°

Then μ=(6cos40°-5)/6sin40°

However, I can't figure out what is the other possible direction.
The one possible is parallel to the wall surface and upward.
what about the other?
Confused.
 
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Hi, yunny.

If you push at the minimum force required to prevent slipping, which way would the picture tend to slide and which way would the force of friction act?

If you push at the maximum force before slipping, which way would the picture tend to slide and which way would the force of friction act?
 
Thanks a lot.

When applying maximum force so that vertical component is greater than the weight, the frame tend to go upward, static friction acts downward.

When applying minimum force, weight is bigger than the vertical component, the frame tends to go downward, static friction acts upward.

problem solved!
 

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