Finding the Minimum Velocity for Safe Rope Sliding: A Physics Problem

  • Thread starter Thread starter Charlie Brown
  • Start date Start date
  • Tags Tags
    Minimum Velocity
AI Thread Summary
To ensure a safe descent on the nylon stockings, the girl must slide down in a controlled manner to keep the tension below 400 Newtons. The minimum velocity for a safe landing depends on the net force acting on her, which is determined by her weight and the tension in the rope. For the elevator scenario, the initial reading of 98 Newtons indicates the force of gravity on the backpack, while the 93 Newtons during motion suggests a downward acceleration. The calculated acceleration of 1.00 m/s² is correct, derived from the difference in forces. The discussion emphasizes the need to calculate the safe sliding method and the minimum velocity required for a safe catch at the bottom.
Charlie Brown
Messages
11
Reaction score
0
Help Please!

I have 2 questions that i need help with:

1) A girl has a mass of 60 kg. She wishes to slide down a rope made of nylon stockings from a second story window. The stockings will break if the tension in then exceeds 400 Newtons. How should she slide down the rope so that she lands safely at a minimum velocity?

The next question I tried...but I'm not sure if i got the right answer.

2) Andrew is on an elevator. Before it begins to move, he pulls a Newton force scale out of his pocket and attaches it to the elevator ceiling. He then hangs his 10.0 kg backpack on the scale and notes a reading of 98 Newtons. When the elevator finally moves the scale adjusts to read 93 Newtons. Find acceleration.

Here's my solution...tell me if its right :biggrin:

Before the elevator moves, the force of gravity acting on the backpack is 98 N...so since there is no motion, the forces are balanced, so the force pulling upwards is also 98 N. When the elevator moves, the scale reads 93 N --> (98 - 5). Since the downward force lost 5 Newtons...I added 5 Newtons to the upward force...so 98 + 5 = 103 N...so if u take up as positive and down as neg. the net force would be 103 - 93 = 10. Since Fnet = ma...you know the mass is 10 kg...so you get 10 = 10a...solving for a you get 1.00 m/s squared. Is that right??
 
Physics news on Phys.org
Use Newton's Third Law: F_{net} = ma

In this case, the tension is the net force on the rope, and you know her mass. So solve for the acceleration.

EDIT: This question asks for the minimum velocity that will land her safely... how fast is safely?
 
Last edited by a moderator:
Jameson said:
Use Newton's Third Law: F_{net} = ma

In this case, the tension is the net force on the rope, and you know her mass. So solve for the acceleration.

EDIT: This question asks for the minimum velocity that will land her safely... how fast is safely?

I'm not sure what the minimum velocity would be...if it helps, there's someone at the bottom of the rope that's supposed to catch her. And the question just asks for the method that she sould used to slide down the rope.
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top