5.25 Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ?(Mass of the man = 65 kg.)

<p>The acceleration of the conveyer belt, a = 1 m/s²</p><p>Coefficient of static friction, &nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mi>s</mi></mrow></mrow></msub></math></span>&nbsp;= 0.2</p><p>Mass of the man, m = 65 kg</p><p>The net force experienced by the man, MA = 1&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mn>65</mn></math></span>&nbsp;N = 65 N</p><p>This net force is due to the friction between the belt and the man</p><p>At maximum static friction</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mi>s</mi></mrow></mrow></msub><mo>×</mo><mi>m</mi><mi>g</mi><mo>=</mo><mi>m</mi><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mi>s</mi></mrow></mrow></msub><mo>×</mo><mi>g</mi></math></span>&nbsp;= 0.2&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mn>10</mn><mo>=</mo><mn>2</mn><mi></mi></math></span> m/s²</p>

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