Using a pulley to raise yourself?

  • Thread starter Thread starter abcdefj
  • Start date Start date
  • Tags Tags
    Pulley
AI Thread Summary
To lift himself steadily upward using a pulley, the man must exert a force equal to the total weight of himself and the platform, which is 950N. The tension in the rope must match this downward force to achieve equilibrium. If he pulls with a force less than 950N, he cannot lift himself, as the tension would not be sufficient. Therefore, he must pull with a force of 950N to raise himself. This confirms that the relationship between the force exerted and the rope tension is crucial for successful lifting.
abcdefj
Messages
3
Reaction score
0

Homework Statement



In this figure, the man and the platform together weigh 950N. The pulley can be modeled as frictionless. Determine how hard the man has to pull on the rope to lift himself steadily upward above the ground. (Or is it impossible? If so, explain why.)

http://www.webassign.net/serpop/p4-33.gif

Homework Equations





The Attempt at a Solution



My thoughts:
a) If there's a 950N force initially downward and he pulls upward (say 475N), wouldn't the tension in the string still be equal to 950N? In this case, wouldn't the man be pulling upward 475N plus an extra 475N? = impossible.

b) The tension in each string equals the total tension in the system (is this true by the way?), so each string would have a tension of 950/2 = 475N, which is how much the man would have to pull upward. = possible.
 
Last edited:
Physics news on Phys.org
What must the tension be in the rope to support the platform plus man?

What is the relationship between the force the man exerts on the rope and the rope tension?
 
Since the man and the platform together weigh 950N (downward force), the tension in the rope must be 950N (upward force).

I'm not quite sure how to go abouts your second question. Isn't the force the man exerts on the rope and the rope tension supposed to be equal?
 
abcdefj said:
Since the man and the platform together weigh 950N (downward force), the tension in the rope must be 950N (upward force).
Good.
I'm not quite sure how to go abouts your second question. Isn't the force the man exerts on the rope and the rope tension supposed to be equal?
Exactly. So how hard does he have to pull?
 
He has to pull 950N. Wow, I was just going in circles there! Thanks :)
 
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...
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}...
Back
Top