Having problems with forces in a static problem

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In a static problem involving forces, tension and friction act to the left, while weight acts downward. For equilibrium, a force must act to the right, which is provided by the normal force, although it is not purely vertical. The normal force has a component that counters torque, ensuring the system remains static. Friction points to the left to maintain equilibrium and prevent rotation, while the normal force's moment is crucial for calculations. Ultimately, the friction coefficient can be determined based on the angle, with a value of approximately 0.29 for 60 degrees.
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I have this situation:

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I know that tension and friction act poining at left. Weight is obviously downward but then I need a force pointing at right for the rod to be static, and the only force left is the normal force but why should it be pointing at left? Isn't it vertical?

Thanks!
 
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why friction is pointing to left if you want equilibrium.
 
No. Normal force has a component to the right. Friction is pointing to left. I think I undeerstand why normal force has a component and it's not vertical only

Thanks!
 
just be careful.normal force is always perpendicular to the direction of motion of point of contact
 
Sorry, I don't get it. What is wrong? Thanks!
 
normal force will be perpendicular to surface ,it does not have left component or right at least in this case.
 
Well then friction should be pointing to left, what are you saying? If friction pointed to right then net torque wouldn't be zero and it would rotate! But if friction pointed to left it would have a translational motion. I need a force pointing to right
 
you will have to take moment about center of mass .normal reaction is opposing that torque.
edit-also if you take moment about point of contact then tension and weight would be counteracting
 
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Ahhhhh now I see it. Normal force has a moment! Thank you very much andrien!
 
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Yeah, that's actually whaat I had to do to calculate the tension force given the fact that I know the weight. After that I can calculate the friction coefficient which doesn't depend on anything, just on the angle theta. It's 0.29 for 60º aproximately
 
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