The force F (in pounds) acting at an angle with the horizontalthat is needed to drag a

Chapter 2, Problem 70

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The force \(F\) (in pounds) acting at an angle \(\theta\) with the horizontal that is needed to drag a crate weighing \(W\) pounds along a horizontal surface at a constant velocity is given by

        \(F=\frac{\mu W}{\cos \theta+\mu \sin \theta}\)

Where \(\mu\) is a constant called the coefficient of sliding friction between the crate and the surface (see the accompanying figure). Suppose that the crate weighs \(150lb\) and that \(\mu=0.3\).

(a) Find \(d F / d \theta\) when \(\theta=30^{\circ}\). Express the answer in units of pounds/degree.

(b) Find \(dF/dt\) when \(\theta=30^{\circ}\) if \(\theta\) is decreasing at the rate of \(0.5^{\circ} / \mathrm{s}\) at this instant.

Equation Transcription:

 

 

 

Text Transcription:

F

theta

W

F=mu W/cos theta+ sin theta

150lb

mu

mu =0.3

dF/d

mu=30 degrees

0.5/s