Force on spinning ice skater's hand as a multiple of hand weight

[Abarak]

Question
A 54.0kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each second. The distance from one hand to the other is 1.5m. Biometric measurements indicate that each hand typically makes up about 1.25% of body weight.

Hand breakdown:
m = .675kg
r = .75m
$$\omega = 2.50\frac{rev}{s} = 15.707\frac{rad}{s}$$
F = 125N

Part A:
What horizontal force must her wrist exert on her hand?
I figured this one out by using $$F=m*r*\omega^2$$

Part B:
Express the force in part (a) as a multiple of the weight of her hand.

My Problem:
I am getting confused with "as a multiple". Pardon my lack of knowledge but I have no idea what they are asking for? I know the answer box is looking for a unit of weight (w).

Sorry about the "easy" question and thank you for taking time in reading my post.

-- Abarak

 

[Dick]

I think you just want to divide the force on the hand by mg. Which would be the weight on the hand. A multiple of weight would not be a weight, but a dimensionless number.

 

[Abarak]

Hey Dick,

Thanks for the help. I was able to get the correct answer.

Man, I could never do what you Homework Helper's do but it REALLY helps!

I am taking an online physics class right now and not having someone to talk to really sucks. If I need to ask my professor a question it takes days to get a response while I can get one here within hours or minutes.

-- Abarak