Minimal coefficient of friction

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Karol
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Homework Statement


The rectangle in the drawing is held by the horizontal rope and the friction on the wall.
What's the minimal coefficient of friction

Homework Equations


Friction force: f=μN

The Attempt at a Solution


Moments round the point of contact with the wall. the side of the rectangle is a, half the diagonal to the center is ##\sqrt {a}##. the distance of the CoM to the wall: ##\sqrt {a}\cdot\sin 55^0=1.16a##
$$1.16a\cdot mg=T\cdot \sin 10^0\rightarrow T=6.68mg$$
$$6.68mg\cdot\mu=mg\rightarrow\mu=0.15$$
It should be 0.3
 
Attachments
  • Free-body diagram of a square tilted 10 degrees, hung from a wall on its left by a horizontal rope at its top-right corner, with a dashed diagonal, angle labels 10 degrees, and a downward weight mg at the centre.
    Snap1.jpg
    11.6 KB · Views: 470
So from the wall to the center of mass is 1.16 times the length of a side ? That's in mid-air ! ;)
 
Is it a square or rectangle? If it's a square check your calculation for "half the diagonal".
 
BvU said:
So from the wall to the center of mass is 1.16 times the length of a side ? That's in mid-air !
I thank you very much BvU for i didn't laugh so much for some time
 
I believe what you used for the normal force is incorrect.
 
The distance of CoM from the wall: ##\frac{a}{\sqrt{2}}\cdot\sin 55^0=0.58a##
$$0.58a\cdot mg=T\cdot \sin 10^0\rightarrow T=3.41mg$$
$$3.41mg\cdot\mu=mg\rightarrow\mu=0.3$$