How Do You Calculate Tension in Ropes on a Frictionless Ramp?

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To calculate the tension in the rope connecting two boxes on a frictionless ramp, the masses are M1 = 0.5 kg and M2 = 2.8 kg, with the ramp inclined at 29°. The tension in the rope between the boxes can be determined using the formula T = M2 * g * sin(29°), where g is the acceleration due to gravity. The user initially thought to use M1 x g x sin(29°) for the second part but realized this was incorrect. The discussion highlights the importance of applying the correct formulas for tension in a system involving inclined planes.
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Two boxes with different masses M1 = 0.5 kg and M2 = 2.8 kg are tied together on a frictionless ramp surface which makes an angle = 29° with the horizontal.

What is the tension in the rope connecting the two boxes? _____N
What is the tension in the rope connecting the first box to the ramp? ______ N

Ok, so I thought for the second part of the problem you would take M1 x g x sin29, however I do not think this is correct. Therefore I am completely stuck. Any help would be greatly appreciated. :)
 
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I just figured it out, sorry for the wasted post.
 
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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