What Causes T2 and T3 to Have No Torque in This Picture?

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T2 and T3 do not produce torque in the given scenario because torque is defined relative to a specific point or axis. In this case, they have no torque about the left end, but may have torque about another point. The system is in equilibrium, meaning the sum of torques around any point must equal zero. However, the concept of torque can still apply to non-equilibrium systems. Understanding the reference point is crucial for analyzing torque in this context.
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In this picture...

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Why do T2 and T3 not produce any torque? Something to do with it being in equilibrium? Thanks.
 
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What point are you taking moments about?
 
Torque in this example has no meaning unless you reference it to a point or axis. T2 and T3 have no torque about the left end, but there may be plenty of torque abot some other point. Although this system is in equilibrium where the sum of torques about any point must sum to 0, the concept of torque about a point applies even for systems that are not 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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