Statics, friction of a rope around a post

AI Thread Summary
The discussion revolves around calculating the maximum tension (T) that a dockworker can support when pulling a rope with a force of 190 N, which wraps around a mooring bit with a friction coefficient of 0.35. Participants clarify that T2 represents the tension in the direction of impending motion, while T1 is the tension exerted by the dockworker. The confusion arises over which tension corresponds to the 190 N force, with consensus that T2 should be treated as the unknown maximum tension. It is emphasized that if T2 exceeds a certain threshold, the rope will slip. The goal is to determine the maximum force of tension before slippage occurs.
J-dizzal
Messages
394
Reaction score
6

Homework Statement


A dockworker adjusts a spring line (rope) which keeps a ship from drifting alongside a wharf. If he exerts a pull of 190 N on the rope, which has 1.25 turns around the mooring bit, what force T can he support? The coefficient of friction between the rope and the cast-steel mooring bit is 0.35.
20150722_171921_zpszgrymvcs.jpg

Homework Equations


T2=T1eμβ
where T2 is in the direction of impending motion.

The Attempt at a Solution


20150722_171849_zpsf7gup5ud.jpg

Not sure what i am missing here, it looks right to me.
 
Physics news on Phys.org
J-dizzal said:
T2=T1eμβ
In this equation, which is the 190N, T1 or T2? Which did you try to solve for?

You should also run through the algebra again.
 
Nathanael said:
In this equation, which is the 190N, T1 or T2? Which did you try to solve for?

You should also run through the algebra again.
T2 is in the direction of impending motion, so i assumed it was 190N. Re reading the problem statement the ship is pulling on the rope with a much greater force, so then T2 would be T, the unknown.
Thank you Nathanael.
 
Last edited:
J-dizzal said:
T2 is in the direction of impending motion, so i assumed it was 190N. Re reading the problem statement the ship is pulling on the rope with a much greater force, so then T2 would be T, the unknown.
If the unknown force of tension were too large, it would cause the rope to slip towards it. We are asked for the maximum force of this unknown tension, so that means the unknown tension is the 'direction of impending motion.'

If you took T2 to be the 190N tension and solved it for T1 then what you would be calculating is the minimum force of tension the other side needs in order for the rope to not slip towards the dockworker.
 
  • Like
Likes J-dizzal
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