<p>The following examples show how to calculate static friction and coefficient of static friction.</p>
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<h2>Example:</h2>
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<h3>Question</h3>
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<p>A block of wood experiences a normal force of \(\text{32}\) \(\text{N}\) from a rough, flat surface. There is a rope tied to the block. The rope is pulled parallel to the surface and the tension (force) in the rope can be increased to \(\text{8}\) \(\text{N}\) before the block starts to slide. Determine the coefficient of static friction.</p>
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<h3>Step 1: Analyse the question and determine what is asked</h3>
<p>The normal force is given (\(\text{32}\) \(\text{N}\)) and we know that the block does not move until the applied force is \(\text{8}\) \(\text{N}\).</p>
<p>We are asked to find the coefficient for static friction \({\mu }_{s}\).</p>
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<h3>Step 2: Find the coefficient of static friction</h3>
<p>\begin{align*} F_f &amp; = \mu_sN \\ \text{8} &amp;= \mu_s(\text{32}) \\ \mu_s &amp;= \text{0,25} \end{align*}</p>
<p>Note that the coefficient of friction does not have a unit as it shows a ratio. The value for the coefficient of friction friction can have any value up to a maximum of \(\text{0.25}\). When a force less than \(\text{8}\) \(\text{N}\) is applied, the coefficient of friction will be less than \(\text{0.25}\).</p>
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<h2>Example:</h2>
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<h3>Question</h3>
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<p>A box resting on an inclined plane experiences a normal force of magnitude \(\text{130}\) \(\text{N}\) and the coefficient of static friction, \(\mu_s\), between the box and the surface is \(\text{0.47}\). What is the maximum static frictional force?</p>
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<h3>Step 1: Maximum static friction</h3>
<p>We know that the relationship between the maximum static friction, \(f_s^{max}\), the coefficient of static friction, \(\mu_s\) and the normal, \(N\), to be:\[f_s^{max} = \mu_sN.\]This does not depend on whether the surface is inclined or not. Changing the inclination of the surface will affect the magnitude of the normal force but the method of determining the frictional force remains the same.</p>
<p>We have been given that \(\mu_s = \text{0.47}\) and \(N=\text{130}\text{ N}\). This is all of the information required to do the calculation.</p>
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<h3>Step 2: Calculate the result</h3>
<p>\begin{align*} f_s^{max} &amp;= \mu_sN \\ &amp;= (\text{0,47})(\text{130}) \\ &amp; = \text{61,1}\text{ N} \end{align*}</p>
<p>The maximum magnitude of static friction is \(\text{61.1}\) \(\text{N}\).</p>
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