5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is

0.15. A force of 200 N is applied horizontally to A. What are

(a) The reaction of the partition

(b) The action-reaction forces between A and B ? What happens when the wall is removed?

Does the answer to (b) change, when the bodies are in motion?

Ignore the difference between μs and μk

 

<p>Mass of the body A, &nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></math></span>&nbsp;= 5 kg</p><p>Mass of the body B, &nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub></math></span>&nbsp;= 10 kg</p><p>Applied force = 200 N</p><p>Coefficient of friction between the bodies and the table, μs = 0.15</p><p><strong> (a)</strong> The frictional force is given by the relation</p><p>fs =&nbsp;<span title="Click to copy mathml"><math><mi>μ</mi></math></span>&nbsp; (&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub></math></span>&nbsp;)g = 0.15&nbsp;<span title="Click to copy mathml"><math><mo>× (</mo><mfenced separators="|"><mrow><mrow><mn>5</mn><mo>+</mo><mn>10)</mn></mrow></mrow></mfenced><mo>=</mo><mi></mi></math></span>&nbsp;22.5 N. towards left</p><p>Hence, the force on the partition = 200 – 22.5 N = 177.5 N, acting rightwards</p><p>According to Newton’s 3rd law, the reaction of the partition will be 177.5 N, acting towards left</p><p>&nbsp;</p><p><strong> (b)</strong> Force of friction on mass A is given by</p><p>Fa =&nbsp;<span title="Click to copy mathml"><math><mi>μ</mi></math></span>&nbsp;mAg = 0.15&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mn>5</mn><mi></mi><mo>×</mo><mn>10</mn><mo>=</mo></math></span>&nbsp;7.5 N, acting leftward</p><p>The net force exerted by mass A on mass B = 200 -7.5 N = 192.5 N, acting rightward</p><p>An equal amount of force will be exerted by mass B on A = 192.5 N, acting leftward</p><p>When the wall is removed, two bodies move in the direction of the applied force</p><p>The net force acting on the moving system = 177.5 N</p><p>The equation of motion for the system of acceleration a, can be written as</p><p>Net force = (&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub></math></span>&nbsp;)a</p><p>a = Net force / (&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub></math></span> )= 177.5 / (5+10) = 11.833 m/s²</p><p>Net force causing mass A to move</p><p>Fa =&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></math></span>&nbsp;a = 5&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mn>11.833</mn><mo>=</mo><mn>59.17</mn><mi></mi><mi>N</mi></math></span></p><p>Net force exerted by mass A on mass B = 192.5 – 59.17 N = 133.33 N</p><p>This force acts in the direction of motion. According to Newton’s 3rd law, force exerted by mass B on mass A will be 133.33 N.</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment