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Chapter 2: Problem 47

Tape is to be coated on both sides with glue by drawing it through a narrow gap. The tape is 0.015 in. thick and 1.00 in. wide. It is centered in the gap with a clearance of 0.012 in. on each side. The glue, of viscosity \(\mu=0.02\) slug \(/(\mathrm{ft} \cdot \mathrm{s})\) completely fills the space between the tape and gap. If the tape can withstand a maximum tensile force of 25 lbf, determine the maximum gap region through which it can be pulled at a speed of 3 ft/s.

Short Answer

Expert verified
The maximum gap region through which the tape can be pulled at a speed of 3 ft/s is approximately 208.33 ft.

Step by step solution

01

Compute the shear stress

We have the glue viscosity \(渭 = 0.02\,\,slug/(\mathrm{ft} \cdot \mathrm{s})\) and the tape speed \(u = 3\,\,ft/s\). The clearance between the tape and the gap is \(y = 0.012\,\,in.\) which is equivalent to \(y=0.001\,\,ft\) as \(1\, in.=0.0833\, ft\). Using the formula for shear stress \(蟿 = 渭 * du/dy = 0.02\,\,slug/ft鈰卻 * 3\,\,ft/s / 0.001\,\,ft = 60\,\,lb/ft虏\) because 1 slug/ft脳s is equivalent to 1 lb鈰卻虏/ft虏.
02

Determine the resistance force

The resistance force is determined using the formula \(F = 蟿 * A\). Here, A denotes the area which the tensile force is acting upon. Since glue is on both sides of the tape, the area becomes double the product of tape breadth and clearance i.e., \(A = 2*b*y = 2*1*0.001 = 0.002\, ft虏\). Now, applying this in the formula for the resistance force \(F = 蟿 * A = 60\, lb/ft虏 * 0.002\, ft虏 = 0.12\, lb\).
03

Calculate the maximum gap region

The maximum tensile force that the tape can withstand is given as 25 lb. Therefore, the maximum gap region that the tape can be pulled through before it breaks is calculated by using the ratio of the maximum tensile force to the resistance force. Hence, Maximum gap region = Maximum Tensile Force / Resistance Force = \(25\, lb / 0.12\, lb/ft = 208.33\, ft.\)

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Most popular questions from this chapter

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