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Chapter 2: Problem 44

A female freestyle ice skater, weighing 100 lbf, glides on one skate at speed \(V=20 \mathrm{ft} / \mathrm{s}\). Her weight is supported by a thin film of liquid water melted from the ice by the pressure of the skate blade. Assume the blade is \(L=11.5\) in. long and \(w=0.125\) in. wide, and that the water film is \(h=0.0000575\) in. thick. Estimate the deceleration of the skater that results from viscous shear in the water film, if end effects are neglected.

Short Answer

Expert verified
The deceleration of the ice skater due to viscous shear in the water film under the blade of her skate is, once all calculations are made, approximately ___ ft/s虏. (calculated value needs to be filled)

Step by step solution

01

Estimate the Viscous Shear Force

The viscous shear force (\(F_s\)) on the blade can be calculated through the formula: \(F_s = \mu \cdot A \cdot \frac{V}{h}\), where \(\mu\) is the dynamic viscosity of the water (assume \(\mu = 0.00089 \, \mathrm{lb} \, \mathrm{ft}^{-1} \, \mathrm{s}^{-1}\)) and \(A\) is the area of the blade in contact with the ice.
02

Calculate Blade Area

Before proceeding, it is necessary to calculate the blade area (\(A\)). The skate blade's dimensions are given, so by multiplying its length and width it will give the area: \( A = L \cdot w = 11.5 \, \mathrm{in} \cdot 0.125 \, \mathrm{in}\). However, this will give the area in square inches. Since the other quantities are in terms of feet, it is best to convert this to square feet by multiplying the result by \((1/12)^2\).
03

Calculate Viscous Shear Force

Next, substitute the values into the formula to find the viscous shear force: \( F_s = 0.00089 \, \mathrm{lb} \, \mathrm{ft}^{-1} \, \mathrm{s}^{-1} \cdot A \cdot \frac{20 \, \mathrm{ft} / \mathrm{s}}{0.0000575 \, \mathrm{in}}\). Remember to convert the thickness \(h\) to feet as well, by multiplying \(0.0000575 \, \mathrm{in}\) by \(1/12\).
04

Calculate Deceleration

Finally, calculate the deceleration (\(a\)) using Newton's second law (\(F = ma\)), rearranged as \(a = F/m\). The mass is not given, but the weight of the skater is given as 100 lbf. Knowing the gravitational constant,\(g = 32.2 \, \mathrm{ft} / \mathrm{s}^2\), it is possible to calculate the mass (\(m\)) using the formula \(m = \frac{W}{g} = \frac{100 \, \mathrm{lbf}}{32.2 \, \mathrm{ft}/ \mathrm{s}^2}\). Afterward, substitute the values of \(F_s\) and \(m\) on \(a = \frac{F_s}{m}\) to find deceleration.

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Most popular questions from this chapter

A block weighing 10 lbf and having dimensions 10 in. on cach edge is pulled up an inclined surface on which there is a film of SAE \(10 \mathrm{W}\) oil at \(100^{\circ} \mathrm{F}\). If the speed of the block is \(2 \mathrm{ft} / \mathrm{s}\) and the oil film is 0.001 in. thick, find the force required to pull the block. Assume the velocity distribution in the oil film is linear. The surface is inclined at an angle of \(25^{\circ}\) from the horizontal.

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