Solve Rope Friction Homework: 55lb Lift 25lb Tire

[aaronfue]

Homework Statement

A 55 lb force is required to lift a 25 lb tire off the ground with a rope wrapped half way around a tree branch. How much force is required to keep a 55 lb girl lifted with the rope wrapped 1.5 wraps around the tree branch?

Homework Equations

T₂ = T₁e^μβ

The Attempt at a Solution

Since I wasn't given a coefficient of friction, my first thought is to figure that out. So far here is how:

T₂ = 55
T₁ = 25
55lb = 25e^μ\pi

By taking the natural log of both sides ln(\frac{55}{25}) = .788
Then dividing that by \pi → μ = 0.251

Now using my coefficient:

T₂ = ?
T₁ = 80
T₂ = 80e^0.251*3\pi

I'm getting an extremely unrealistic number for this, so I'm not sure where my errors are.

 

[Chestermiller]

Homework Statement

A 55 lb force is required to lift a 25 lb tire off the ground with a rope wrapped half way around a tree branch. How much force is required to keep a 55 lb girl lifted with the rope wrapped 1.5 wraps around the tree branch?

Homework Equations

T₂ = T₁e^μβ

The Attempt at a Solution

Since I wasn't given a coefficient of friction, my first thought is to figure that out. So far here is how:

T₂ = 55
T₁ = 25
55lb = 25e^μ\pi

By taking the natural log of both sides ln(\frac{55}{25}) = .788
Then dividing that by \pi → μ = 0.251

Now using my coefficient:

T₂ = ?
T₁ = 80
T₂ = 80e^0.251*3\pi

I'm getting an extremely unrealistic number for this, so I'm not sure where my errors are.

Where did the 80 come from. Shouldn't it be 55 lb? Also, what unrealistic number did you get?

 

[aaronfue]

Where did the 80 come from. Shouldn't it be 55 lb? Also, what unrealistic number did you get?

Well the 80 came from the weight of the tire combined with the girl. 25lb + 55lb

I ended up, after putting it all in my calculator a few times, with a force of 900+. As I said, a completely unrealistic number to hold up a child+tire.

 

[Chestermiller]

In the case of lifting the tire, the 55 lb force has to overcome the coefficient of friction in the rotational direction oriented from the 25 lb tire to 55 lb force. In the case of keeping the girl raised, the frictional force is acting in the opposite direction to keep the girl and tire from slipping downward. 80 exp(-3π(0.251)) = 7.5 lb.