# The USS Constitution’s largest anchor weighed 5443 lbs

1. Oct 17, 2003

### tucky

Here is two questions about torque that I am having problem figuring out:

Q: The USS Constitution’s largest anchor weighed 5443 lbs. The capstan was a drum that measured roughly 4 ft in diameter. Twelve poles, each about 12 feet long, could be stuck into the capstan. Sailors pushed on the poles, winding the anchor rope around the capstan and raising the anchor. Assume half the torque generated is lost to friction and slippage. How much force must be applied to each pole to raise the anchor? How many sailors do you think would have to man each pole?

A: Torque = r * F
Torque= 14ft *5443lb = 76202ft-lb multiply it by two because half of the torque is lost = 152404ft-lb /12 poles =12300.3ft-lb. –that just seems to high

Q: A kid turns his bike upside down and spins the tire. The spoked wheel measures 26 inches in diameter and weighs 5 lbs. The kid presses a piece of metal against the rubber tire to bring it to a halt. About how long will it take to bring the wheel to a halt if the kid presses with a force of 10 lb?

A: I am not really sure how to work this problem. I think this is a Newton 2 law problem: Sum of forces = ma
10lb = 4.54 kg * 9.8N =44.49N
I = ½ 2.27(.3302m)^2 = .1237kgm^2
44.49N=.1237kgm^2 * a/.3202m if I solve it this way I get a –115.161/s^2 acceleration which is wrong. I really do not know what to do?

Last edited by a moderator: Feb 6, 2013
2. Oct 17, 2003

### HallsofIvy

Staff Emeritus
Why? It's not the anchor that is pushing against the capstan poles!

The weight of the anchor is 5443 lb and the anchor rope is taken up around the capstan which is 2 ft in radius- the torque neccessary to do that is 5443*2. There are 2 poles of length 12 ft (you are assuming that that is in addition to the 2 ft of the capstan- ok) so if F is the force applied, the torque due to each one is 14F. The torque due to both is 2*14F but we are told that half the torque is "lost to friction and slippage" so that puts us back to 14F. We must have 14F= 5442*2 which we can solve for F.

How much force do you think each sailor can apply?

I have no idea how to do this problem. We are told the weight of the wheel but are not told how that weight is distributed about the wheel which is critical. Also knowing how much force is exerted against the tire doesn't tell us what the friction force is.

3. Oct 18, 2003

### tucky

HallsofIvy,

Thank you!!!!! I agree about the wheel problem. Do you think most of the weight is distributed like a hoop? Even assuming that, I still think that something is missing from the question. Thanks again!

4. Oct 18, 2003

### gnome

For the first question, there are 12 poles, not 2. Also, I'll agree it's arguable, but I would have assumed that the poles would stick pretty far into the capstan to avoid the risk of slipping out. And the sailors wouldn't be pushing exactly at the tip of the poles. So I would use just 12 feet as the effective length for computing the torque.
So that would give the force on each pole as
f = 5443*2*2/(12*12) = 151 lb. (the extra 2 in the numerator accounts for the friction).
One sailor per pole might do the trick, but with less than perfect traction on the deck, 2 or 3 would probably be better.

maybe as little as .2 s, or as much as .6 s, but what's a few tenths of a second?

The question says ABOUT, so be creative. You're missing a few details: the coefficient of friction, the moment of inertia, and the speed at which he spins the wheel, but you can estimate them.
We know:
N (normal force) = 10 lb
r = 13/12 ft
m = 5/32 slug

My textbook says that the coeff. of kinetic friction for steel on steel is .57, and for rubber on concrete it's .8. So I'll take
mu;k = .7 as a reasonable guess.

Next, a bicycle wheel is somewhere between a uniform solid cylinder and a uniform hoop, which makes its moment of inertia I:
(1/2)mr2 <= I <= mr2
I'll assert that its probably closer to the hoop, so I'm using mr2 as my estimate, but I'll keep in mind that it's actually maybe .8 or .9 times that. So,
I = (5/32)(13/12)2 slug*ft^2 (yeccch)

The torque he applies (as a result of the friction): T = -&mu;N*r
T = -.7*10*13/12

Since T = I&alpha;, the angular acceleration:
&alpha; = T/I = -(.7*10*13/12)/(5/32)(13/12)2

Finally, the initial speed. If the boy is REALLY energetic, MAYBE he can get it going close to 20 mi/h. If he's a wus, only 5 mi/h. I'm going to use 10 mi/h.
v = 5280*10/3600 ft/s but we need angular velocity so thats &omega; = v/r
&omega;i = 5280*10*12/(3600*13)

So, wrapping it all up, &omega;f = &omega;i + &alpha;t , and &omega;f=0 , therefore ...
t = -&omega;i/&alpha;
t = [-5280*10*12/(3600*13)] / [-(.7*10*13/12)/(5/32)(13/12)2]
t = (52800*5/32)/(3600*7)
t = approx. .3 s

t is directly proportional to v and I, and inversely proportional to &mu;k
Maybe I'm off by 10% for &mu;, maybe I'm off by 10% or even 20% for I, maybe I'm off by 50% for vi. So what? The question was ABOUT how long. For sure, it's more than .1 sec, and less than .8 sec.

5. Oct 18, 2003

### tucky

gnome,

Thank you! That makes since to me and you made me laugh!