How Do You Analyze Forces in a Pulley System Equilibrium Problem?

AI Thread Summary
To analyze forces in a pulley system at equilibrium, the key equations are the sum of torques and forces, which must equal zero. The tension in the upper cable is denoted as T, with the weight of the hanging block represented as Mg. Each segment of the cable experiences different tensions due to the arrangement of the pulleys, specifically T/2 for the top pulley and T/4 for the middle pulley. The total tension from the segments must balance the weight, leading to the equation 7T/8 = Mg. The solution reveals that the tension can be calculated as T = 44.8.
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Homework Statement


The picture is located in the attachment

Homework Equations



\sum t = 0

\sum Fx = 0 Although there are no horiz. forces in this problem

\sum Fy = 0

The Attempt at a Solution



I'm not sure how to start. On the top most pulley, I believe that there is a Tension T and a force Mg that is from the hanging block. The second pulley has a Tension pulling it up and Mg pulling it down. The last pulley has a Tension pulling it up and Mg as well as F pulling it down. First, did I analyze the forces correctly?

Or Does the entire system have a Tension pulling it up, a Force F pulling it down, and 3Mg pulling it down as well?
I'm quite confused.
 

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If T is the tension in the upper cable, the tension in the each segment of the cable passing through the top pulley will be T/2.
Tension in the each segment of the cable passing through the middle pulley will be T/4.
Tension in the each segment of the cable passing through the bottom pulley will be ...
Total tension in the three segments of the three cables passing through the pulleys balances the weight.
Equate them and find the tension T.
 
So the three tensions acting on the block is T/2, T/4, T/8.

7 T/8 = Mg

T = 44.8

How did you figure out that those tensions in the cable (the one passing the middle pulley being T/4 and the one passing the bottom pulley T/8)?
 
In a single string, each segment has the same tension if the string is in equilibrium.
So T in the top cable is due to T/2 in each segment of the cable passing through the pulley.
 
Okay. Thanks
 
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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