Calculating Tension Forces in a Frictionless Pulley System

[Destrio]

A man standing on a platform with an attached frictionless pulley is supporting himself at a fixed position in midair (zero acceleration). What force F is required to do this? Solve for F for a man of arbitrary mass M, and a platform plus pulley of arbitrary mass m.

So, since a = 0 : Fnet = 0
Fnet = Fu - Fg = 0
Fu = Fg
Fu being the upward force exerted by the man

I'm not sure if I can use Fu as an upward force since there will be 2 tensional forces

How can I apply the T forces between the man and the pulley, and the pulley and the roof
could i do
Fnet = T1 + T2 - Fg = 0
?

Thanks

 

[Doc Al]

You're on the right track. Consider the man+platform as a system. What forces act on it?

Hint: What can you say about the tension in each section of the rope?

Hint 2: How does the force the man exerts relate to the tension in the rope?

 

[Destrio]

for the man + platform we have force of gravity, the normal force of the platform on the man (although that could be irrelevant if we are thinking of the man+platform as a system), and the tensional force of the string the man is holding

would the tension of each section of the rope be equal?
t1 = t2 ?

the force the man exerts should be equal to the tension of the rope?

 

[Doc Al]

All correct. Keep going.

 

[Destrio]

ok
Fnet = 0
t1 = t2
Ttotal = t1 + t2 = Fman

im not sure how to relate the force of gravity to these other forces

would the force of gravity be opposite to the total tension forces, being equal to the force the man exerts

or, would the force the man exerts be equal to a single tension force:
t1 = t2 = Fman

making the force the man exerts twice the force of gravity?

 

[Doc Al]

ok
Fnet = 0
t1 = t2

OK.

Ttotal = t1 + t2 = Fman

Not OK.

im not sure how to relate the force of gravity to these other forces

would the force of gravity be opposite to the total tension forces,

Yes!

being equal to the force the man exerts

No!

or, would the force the man exerts be equal to a single tension force:
t1 = t2 = Fman

Yes.

making the force the man exerts twice the force of gravity?

No.

(1) What upward forces act on the system?
(2) What downward forces act on the system?

 

[Destrio]

ok
Fnet = 0
t1 = t2 = Fman

t1 + t2 = -Fg
is this correct?

upward forces:
Force of the man on the string
Force of the roof on the string

downward forces:
force of gravity on the platform + man
force of the string on the man
force of the string on the roof

Fms + Frs = Fg + Fsm + Fsr

Fms = Fsm
Frs = Fsr

Do I have to account for any normal forces?

 

[Doc Al]

Fnet = 0
t1 = t2 = Fman

OK.

t1 + t2 = -Fg
is this correct?

Almost. From Fnet = 0 we know that: t1 + t2 -Fg = 0
Or: t1 + t2 = Fg

That's really all you need to solve this one. (Combine these equations together to solve for Fman.)

upward forces:
Force of the man on the string
Force of the roof on the string

The upward force on the system is just the pull of the rope. The rope attaches twice--once to the man's hand and once to the platform--so the total upward force is twice the tension.

downward forces:
force of gravity on the platform + man
force of the string on the man
force of the string on the roof

The downward force is just the weight of the system.

 

[Destrio]

Fnet = 0
t1 = t2 = Fman
t1 + t2 - Fg = 0
t1 + t2 = Fg

so

Fman + Fman = Fg
2Fman = Fg
Fman = (1/2)Fg

 

[Doc Al]

Perfecto! Now just express Fg (and then Fman) in terms of the given information: M and m.

 

[Destrio]

ah yes

Fnet = 0
t1 = t2 = Fman
t1 + t2 - Fg = 0
t1 + t2 = Fg

Fman + Fman = Fg
2Fman = Fg
Fman = (1/2)Fg

Fman = (M+m)a
Fg = (M+m)g

Fman = (1/2)(M+m)g

 

[Doc Al]

Excellent.

 

[Destrio]

thanks very much for your help