How Do You Calculate the Tension in a String With Hanging Mass and Pulleys?

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To calculate the tension in the string with a 12 kg hanging mass in equilibrium, one must consider the forces acting on the system. The tension must balance the weight of the mass, which is 12 kg multiplied by the acceleration due to gravity, resulting in a force of 117.6 N. The angles of the string segments complicate the calculations, as they are not provided, making it difficult to apply trigonometric relationships. Attempts to use wave properties or frequency to find tension were unsuccessful due to a lack of necessary parameters. Ultimately, the tension in the string can be simplified to match the weight of the hanging mass when ignoring the mass of the string itself.
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Homework Statement



A 12 kg object hangs in equilibrium from a string with a total length of L=5m and a linear mass density of u= 0.001kg/m. The string is wrapped around two light, frictionless pulleys that are separated by a distance of d=2m. Determine the tension in the string.

Sorry i don't have a picture but - the pulleys are at the top an the rope goes around them with a hanging mass, forming an inverted triangle.

Homework Equations





The Attempt at a Solution



Normally, i would write equation for the sum of the forces on the x and y axis:

If there was an axis at the bottom of the inverted triangle and the angles were measured from the axis to the strings;

sum of forces on x-axis = T(right) cos theta - T(left) cos theta = 0
sum of forces on y-axis = mg - T(right)sin theta - T(left)sin theta = 0,
But since i don't know the angles, this goes nowhere.

I have also tried using;
v = Square root of (tension/u), but I don't have v,
So i tried to find v using f = v/ wavelength, but i don't have frequency.

I have tried so many things and I can't seem to get a start. Any hints?
 
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If one ignores the mass of the string the tension in it is simply the weight of the 12 kg object. This one can deduce from the fact that the tension in the string needs to balance the weight of the object in order to keep it 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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