Free-Body Diagrams: Solving for Tension and Weight

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To draw a free-body diagram accurately, it's essential to represent all forces acting on the object, including weight (W), tension (T), and the normal force (N). The treatment of the object as a particle or one with dimensions affects the diagram's correctness, as two of the three proposed diagrams may be valid based on this consideration. When the object has dimensions, summing torques about a specific point, like the lower right corner of the box, is necessary to ensure stability and rotational equilibrium. Understanding these principles is crucial for correctly analyzing forces in physics problems. Accurate free-body diagrams are fundamental for solving tension and weight issues effectively.
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free-body diagram (urgent)

I was wondering how to draw a free body diagram, could u see if this is correct? A body of weight W was pulled along a rough surface, with a string and applied tension T, which one of the following diagrams is correct? N is normal contact force.
 

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Sometimes objects are treated as particles, and at other times, objects are considered to have dimensions. So 2 of the three could be correct, depending on whether you consider the object to have dimensions. For the latter case, you'll need to sum torques about the lower right hand corner of the box to check stability and compatability with rotational 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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