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Chapter 1: Problem 76

The sled shown in Fig. \(P 1.76\) slides along on a thin horizontal layer of water between the ice and the runners. The horizontal force that the water puts on the runners is equal to \(1.2 \mathrm{lb}\) when the sled's speed is \(50 \mathrm{ft} / \mathrm{s}\). The total area of both runners in contact with the water is \(0.08 \mathrm{ft}^{2}\), and the viscosity of the water is \(3.5 \times 10^{-5} \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}\) Determine the thickness of the water layer under the runners. Assume a linear velocity distribution in the water layer.

Short Answer

Expert verified
The thickness of the water layer under the runners is calculated by manipulating the equation for the viscous force.

Step by step solution

01

Understanding given variables

Let's define the given variables: The viscosity of the water (η) is 3.5 x 10^-5 lb·s / ft^2, the total area of contact (A) is 0.08 ft^2, the force of the water on the runners (F) is 1.2 lb, and the sled's speed (V) is 50 ft / s.
02

Formulate the equation for the thickness of the fluid using the known variables

The thickness of the fluid (d) is found by rearranging the formula for the viscous force, which gives: d = (ηA x V) / F.
03

Substitute the given values into the equation

Substituting the known values, we get: d = ((3.5 × 10^-5 lb·s / ft^2) x (0.08 ft^2) x (50 ft / s)) / 1.2 lb.
04

Compute the value of d

Solving the equation yields the thickness of the water layer under the runners.

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Most popular questions from this chapter

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