Calculating Force of Falling Weight in Closed System

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In a closed system involving a cart and falling weights, the force of the falling weight is calculated using the formula mass times acceleration, where acceleration is derived from the cart's speed over a distance. It is incorrect to simply use mass times gravity when the weight is connected to a pulley, as both gravitational force and tension in the rope must be considered. The net force acting on the falling weight is the sum of gravitational force and tension, while tension cancels out when analyzing the entire system. For the overall system, the net force is the combined mass of the weights multiplied by acceleration, equating to the gravitational force plus any additional forces acting on the system. This approach ensures accurate calculations of forces in a closed system setup.
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I'm just wondering if I'm doing this correctly..

If I have something like this: http://img326.imageshack.us/my.php?image=pull6cx.png

It's a closed system where the weights on the cart will get placed on the bottom holder for each trial... To find the 'force of the falling weight' I would take the mass*acceleration, where acceleration is determined by how fast the cart goes in a certain distance right? I wouldn't just take the mass*gravity when it's connected to a pulley like that would I?
 
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The only forces acting on the falling weight are Fg1 (force of gravity) and FT (force of tension in the rope). For the falling weight alone, Fnet1 = ma = Fg1+FT.

Since FT is applied in both directions in the rope, it "cancels out" when you involve the entire system. For the entire system, Fnet = (m1+m2)a = Fg1+Ff2.

Hope that helps.
 
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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