How Do You Calculate Tension in Strings Supporting a Meter Stick in Equilibrium?

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To calculate the tension in the strings supporting a uniform meter stick in equilibrium, first convert the mass of the stick from grams to kilograms, resulting in 0.18 kg. Using the formula F=mg, the weight of the meter stick is determined to be approximately 1.764 N. The problem requires applying the equilibrium condition that the sum of vertical forces equals zero. The tensions in the strings at the 0 cm and 90 cm marks must balance the weight of the stick while considering the moments about a pivot point. A clear understanding of the forces and moments involved is essential to solve for the tensions accurately.
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Homework Statement


A uniform meter stick with a mass of 180 g is supported horizontally by two vertical strings, one at the 0 cm mark and the other at the 90 cm mark. What is the tension in the string at 0 cm and at 90 cm?



2. The attempt at a solution I changed 180 g to .18 kg. Using F=mg, I found mg=1.764 N. From there, I have made numerous attempts, but none make sense! I am making this problem way harder than it should be. Please help!
 
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Sorry, I also know that the sum of the forces needs to be 0; and there is only forces in the y direction.
 
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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