Finding the direction of forces

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To accurately represent tension in a free body diagram (FBD), place the tail of the tension arrow at the center of mass if treating the object as a point mass. For extended bodies, position the tail at the attachment point of the string or rope. The arrow's shaft should be parallel to the taut string, with the head pointing away from the object, as ropes cannot push. If further clarification is needed, participants are encouraged to ask additional questions. Understanding these principles is crucial for correctly analyzing forces in physics.
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Homework Statement
how can i easily find the forces direction like if i said a kid was pulling 10lb[W] where would i put the tension in the free body diagram
Relevant Equations
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Tension force = 10lb[W] where do i put tension arrow in the full body diagram
 
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You're welcome. After reading through that article, if you still have questions please post them here and we will try to help you understand FBDs.
 
To draw an arrow representing tension:
  • If you treat the system as a point mass, put the tail of the arrow roughly at the center of mass of the object. If it is an extended body and you need to calculate torques, then it would be correct to put the tail of the arrow at the point where the string or rope is attached to the object.
  • The shaft of the arrow should be drawn parallel to the taut string or rope.
  • The head of the arrow should always point away from the object. That's because you cannot push with a rope.
I hope this helps.
 
On the free body diagram of what body?
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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