Attrito nelle funi

ITIS OMAR Dipartimento di Meccanica
Attrito nelle funi
Problema n. 1
Determine the force P required to (a) raise and (b) lover the 40-kg cylinder at a slow steady speed. The
coefficient of friction f between the cord and its supporting surface is 0.30
L'ampiezza α dell'arco di contatto vale:
α = π rad
(a) La forza P necessaria per far salire senza accelerazione il corpo vincolato dalla fune vale:
P
= exp (α f ) = 2.566
mg
P = mg ⋅ exp (α f ) = 1007 N
(b) La forza P necessaria per far scendere senza accelerazione il corpo vincolato dalla fune vale:
mg
= exp (α f )
P
mg
P=
= 152.9 N
exp (α f )
1
ITIS OMAR Dipartimento di Meccanica
Problema n. 2
A force P=mg/6 is required to lower the cylinder at a constant slow speed with the cord making 5/4 turns
around the fixed shaft. Calculate the coefficient of friction f between the cord and the shaft.
mg
5

= exp  π ⋅ f 
mg
2

6
5
6 = exp  π ⋅ f 
2

5
log ( 6 ) = π ⋅ f
2
2log ( 6 )
f =
= 0.228
5π
2
ITIS OMAR Dipartimento di Meccanica
Problema n. 3
A dockworker adjust a spring line (rope) which keeps a ship from drifting alongside a wharf. If he exerts a
pull F of 200 N on the rope, which has 5/4 turns around the mooring bit, what force T can be support? The
coefficient of friction f between the rope and the cast.steel mooring bit is 0.30
T
5
= exp  π ⋅
F
2

f

5

T = F ⋅ exp  π ⋅ f 
2

5

T = 200 ⋅ exp  π ⋅ 0.3 
2

T = 2110 N
Problema n. 4
If the dockworker of Prob.3 is to support a tension T of 18 kN in the rope leading to the ship, how many
turns around the mooring bit are necessary if he exerts a pull F of 240 N on the free end? The coefficient of
friction f between the rope and the bit is 0.30
T
= exp (α f )
F
T 
α f = log  
F
T
log  
 F  = 75 = 250 rad
α=
f
0.3
250
n=
= 39.7 giri
2π
Per assicurare l'ortogonalità tra T ed F il numero minimo di giri di avvolgimento sarà pari a 40,25
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ITIS OMAR Dipartimento di Meccanica
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Problema n. 5
In western movies, cowboys are frequently observed hitching their horses by casually winding a few turns of
the reins around a horizontal pole and letting the end hang free as shown. If the freely hanging length of rein
has a mass m of 0.060 kg and the number of turns is as shown, what tension T does the horse have to
produce in the direction shown in order to gain freedom? The coefficient of friction f between the reins and
wooden pole is 0.70
L'angolo di avvolgimento α della fune sul palo vale:
π
π 13
α =  2π −  + 2 π + = π
6
2 3

La forza T necessaria alla liberazione del cavallo vale:
T
 13
= exp  π ⋅
mg
 3

f

 13
T = mg ⋅ exp  π ⋅ f
 3

 = 0.5886 ⋅ 13759.68 = 8099 N

ITIS OMAR Dipartimento di Meccanica
5
Problema n. 6
For a certain coefficient of friction f and a certain angle α , the force P required to raise m is 4 kN and that
(F) required to lower m at a constant slow speed is 1.6 kN. Calculate the mass m.
 P
 π

 mg = exp   2 + α 




 mg
 π

= exp   + α 

 2
 F
 mg
1
f ⇒
=
P
 π

exp   + α 

 2

f

m2 g 2
=1
F⋅P
F⋅P
m=
= 257.88 kg
g

f

ITIS OMAR Dipartimento di Meccanica
Problema n. 7
A 50-kg package is attached to a rope which passes over an irregularly shaped boulder with uniform surface
texture. If a downward force P = 70 N is required to lower the package at a constant rate, (a) determine the
coefficient of friction f between the rope and the boulder. (b) What force F would be required to raise the
package at a constant rate?
(a) Poiché il peso del pacco e il tiro della fune hanno direzione comune, indipendentemente dall'irregolarità
del masso, l'angolo di avvolgimento α vale π .
mg
= exp (π f )
P
 mg 
π f = log 

 P 
 mg 
log 

 P  = 1.947 ≅ 0.62
f =
π
π
(b)
F
= exp ( π f )
mg
F = mg ⋅ exp (π f ) = 490.5 ⋅ 7.013 ≅ 3440 N
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ITIS OMAR Dipartimento di Meccanica
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Problema n. 8
The 80-kg tree surgeon lowers himself with the rope over a horizontal limb of the tree. If the coefficient of
friction f between the rope and the limb is 0.60, compute the force F which the man must exert on the rope to
let himself down slowly.
L'angolo di avvolgimento α vale π
mg
= exp (α f )
F
mg
= exp ( π f )
F
mg
784.8
F=
=
= 119.2 N
exp ( π f ) 6.586