Gioncoli 4th ed chap4 q59, tension

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The discussion revolves around a physics homework problem involving tension in a system with two masses. The user initially assumes no vertical acceleration and sets up equations based on tension, leading to relationships between the angles and accelerations of the masses. They struggle to eliminate the variable aC from their equations but later realize that aC is equivalent to ax, simplifying their approach. This realization helps clarify their understanding of the problem. The conversation highlights the importance of careful reading and comprehension in solving physics problems.
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Homework Statement



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Homework Equations


The Attempt at a Solution



I assumed to meet the condition, there can be no vertical acceleration so if i assume the string of mass mB to make an angle theta with the vertical, then if T is tension,

Tcos(theta) = mBg ...1
Tsin(theta) =mBax...2Since the 2 masses are relatively at rest then their acceleration (horizontal ) should be equal

so T = mAax...3

eq 2\3 gives sin(theta) = mB\mA

solving eq 3 and 1 i get ax=gtan(theta)

also

mcac=F-T-Tsin(theta)

and mCaC=F-(mA+mB)ax

But I can't find a way to get rid of aC

Is there something wrong with my approach or reasoning?
 
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Hi Idoubt! :smile:
Idoubt said:
But I can't find a way to get rid of aC

No, that's fine …

aC is aX :wink:​
 
mA does not move relative to mC, so aC = ax .
 
Doh ! *slaps myself*... *twice*...*moans in agony*...

gotta remember to read questions properly before diving in :blushing:

thx guys
 
Last edited:
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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