Solving a Mechanics Problem: Equilibrium and Tension in a 160kg Box

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To maintain equilibrium for a 160kg box, the tension must equal the weight of the box, which is 1600N. The discussion emphasizes that with only two forces acting on the box, the tension must balance the weight for the system to remain stable. Participants express difficulty viewing the accompanying diagram but agree on the fundamental principle that the opposing force must match the weight for equilibrium. The conversation highlights the importance of understanding free body diagrams in mechanics. Overall, the key takeaway is that tension must equal the weight of the box to achieve equilibrium.
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http://hk.pg.photos.yahoo.com/ph/good_students/detail?.dir=7712&.dnm=7c7f.jpg

What should be the value of P to maintain the system in equilibrium?
If we just consider the free body diagram of the 160kg box, should the tension be 1600N?(Because there are only 2 forces acting on the box only)
Could any people please help?
 
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Sorry mate...couldn't see your diagram (it doesn't show up).
 
I could not see it either, but according to what you said, you are correct:

IF there are only two forces on an object, and one of them is weight, then the other must be equal to the weight if the object is in equilibrium.
 
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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