How Does Tension Affect the Acceleration of Blocks in a Pulley System?

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The discussion focuses on a physics problem involving a pulley system with two blocks, where one block weighs 486 N and the other weighs 190 N. Participants note the difficulty in solving the problem without a diagram, emphasizing the importance of visual aids in understanding the forces at play. The main objectives are to calculate the acceleration of the blocks and the tension in the cord, assuming no friction and a massless pulley. The conversation highlights the need for clarity and calmness when tackling complex physics problems. Ultimately, the thread underscores the collaborative nature of problem-solving in physics.
charvonne coates
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By the grace of god please help!

tension problem?


[CJ6 4.P.067.] In the drawing, the weight of the block on the table is 486 N due east and that of the hanging block is 190 N due south. Ignore all frictional effects, and assume the pulley to be massless.


(a) Find the acceleration of the two blocks.

(b) Find the tension in the cord


:cry:
 
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without the diagram, it is quite hard for us to solve it.
 


It appears that you may have accidentally posted your physics problem in the wrong forum. However, by the grace of God, I hope that you find the guidance and assistance you need to solve this problem. Remember to approach it with a clear and calm mind, and trust in your abilities to find the solution. Best of luck to you.
 
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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