Physics Tension Force Homework Problem

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A ball of mass m is suspended by two strings of unequal lengths, leading to different tension forces T1 and T2. The relationship between these tensions is established as T1<T2, with T1 sloping downwards and T2 hanging nearly straight down. Since the ball is in static equilibrium, the horizontal components of the tensions must be equal, indicating no net horizontal force. This understanding helps clarify the relative proportions of the tensions involved. The problem emphasizes the importance of analyzing forces in equilibrium to solve for tension relationships.
dlilpyro
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I need some help with this problem. A ball of mass m is suspended from 2 strings of unequal length. The tensions in T1 and T2 in the strings must satisfy which of the following relations? The answer is T1<T2. I can't get a lucite in but T1 is on the left and sloping down to the right on an incline while T2 is nearly straight down. U2 is the long string and T2 is the short string.
 
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Try equating the horizontal components of the tension. You know that the object is in static equilibrium and so there is no horizontal force on it. Working backwards from the fact that the horizontal components must be equal will give you an idea for the relative proportions of the tensions.
 
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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