Tension Troubles: Understanding Mass and Force

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Tension in a hanging string with equal masses on both ends is indeed equal to the weight of one mass times gravity, confirming that T equals 500N if each mass is 500N. When masses of different weights are involved, the tension in the string corresponds to the weight of the lighter mass. Additionally, when a string is pulled in opposite directions with equal force, the tension is equal to the force of one side, not the sum of both forces. Understanding these principles can be clarified through free body diagrams, which visually represent the balance of forces. This knowledge solidifies the foundational concepts of mass and force in physics.
catenn
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Hey, I have heard a few things about tension I am unsure about and would like to check. Is it true if there is a string hanging with both ends and the same mass on each the tension is equal to the mass of one of the weights times gravity? (ex. on each end of hanging string and pulley 500N, so T=500N?) And then if there are masses of different weights the Tension is of the lighter mass?

Also, if there were to be one string being pulled in opposite directions with the same force, the force is of one of them and not the sum of the two forces?

Thanks!
 
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The answer to your first question is yes. You can best show it by drawing a free body diagram of the system -- that shows how the tension in the string balances each mass' weight. Here is a PF thread that may help to clarify the situation for you:

https://www.physicsforums.com/showthread.php?t=130847
 
Okay! Thanks, finally a click! Things make sense! :) I have been trying to put together different parts and see it now.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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