Reaction and friction coefficient on a cylinder

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SUMMARY

The discussion focuses on calculating the tension in a rope holding a cylinder weighing 200 N on a rough surface and determining the minimal coefficient of friction required. The participant initially calculated the tension as 66.7 N but later corrected it to 654 N after realizing the weight should be considered in terms of mass (200 kg). The torque balance equation used was T = 200R·sin(30°)/d, where d was derived from the geometry of the setup.

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Karol
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Homework Statement


The cylinder weighs 200[N] and lies on a rough surface. a rope holds it in place.
What's the tension T in the rope and what's the minimal coefficient of friction.

Homework Equations


Friction: f=μN

The Attempt at a Solution


All the geometry i have drawn. T balances the torque made by the weight of the cylinder. i calculate torque round the point of contact with the inclined surface:
The distance d from the line of action of T:
$$d=\frac{R}{\tan30^0}\cdot \sin 60^0$$
Now the torque from the weight equals the torque from T:
$$200R\cdot \sin 30^0=T\cdot d\rightarrow T=66.7$$
It should be 654 and i don't continue yet to the friction coefficient.
Edit: i found my mistake, in Newtons it comes out right. now i will try to find the coefficient of friction
 

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I get 654 N if the cylinder has a mass of 200 kg instead of a weight of 200N.

654/66.7=9.81 made the difference clear.
 

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