Will the Rope Hold the Archaeologist Safely?

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The discussion centers on the tension in a rope supporting an archaeologist crossing between two cliffs. The rope can withstand a maximum tension of 2.95 x 10^4 N, while the archaeologist's mass is 93.8 kg. Incorrect calculations previously suggested a tension of 4817.6 N and a minimum angle of 1.78 degrees. The correct approach involves evenly distributing the archaeologist's weight, resulting in a tension of approximately 2408.8 N and a minimum angle of 0.89 degrees. Properly accounting for the weight distribution is crucial to ensure the rope does not break.
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An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope . The rope will break if the tension in it exceeds 2.95EE4 (2.95 x 10^4) , and our hero's mass is 93.8 .

<img src=http://session.masteringphysics.com/problemAsset/1041292/5/YF-05-42.jpg>

A. If the angle between the rope and the horizontal is = 11.0, find the tension in the rope.
B. What is the smallest value the angle can have if the rope is not to break?

The following answers are incorrect:
A. 4817.6 N
B. 1.78 degrees
 
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Do you have your calculations at all?

Also the link you have posted is broken.
 
You did what many people get wrong. You forget to evenly distrubute the weight of the person to both sides of the rope. Since he is in the exact middle then you will divide his weight in half and put them on both the "y" dimension.

From this I got a) 2408.796 N
b) .89 degrees

Notice these are exactly half of your answer this is because I halved the weight of each side.
 
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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