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Static and kinetic friction

Created By: Shay Meni
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Question:
A box is lying still on an inclined plane. The inclination angle (from the horizontal) of the plane is $alpha=30^circ$ . At a certain moment, the box is being punched. This punch sets the box in motion up the inclined plane. The box climbs until it comes to a stop. The scenario is depicted in the clip. The coefficient of kinetic friction between the box and the plane is $mu_k=0.5$ What is the lower limit that can be placed on the value of the static friction coefficient, $large mu_s$ ?

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True. If the box can lie still, it must be that the the static friction force, $f_s$ , and the component of the weight balance, so $f_s=mg m{sin}(alpha)$ . In addition, $f_sleq mu_s N$ . The normal force balace the prependicular component of the weight, therefore $N=mg m{cos}(alpha)$ . combining these three relation we get $m{tan}(alpha)<mu_s$

Not true. The box will come to a stop and then remain at rest due to static friction force.

Wrong. It might be that the static friction coeffcient is larger that 0.5 but a stronger lower limit can be placed.

Wrong for many reasons. 1. This is not a lower limit. 2. This is not true. it is possible to have $mu>1$

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Arraba. george shchupak Created At: 2017-06-24 21:58:27

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History of edits

Edited BY: Meni Shay Edit Date: 2017-05-22 17:53:43

Edited BY: Meni Shay Edit Date: 2017-05-22 17:50:57

Edited BY: Meni Shay Edit Date: 2017-05-22 17:40:30

Edited BY: Meni Shay Edit Date: 2017-05-22 17:37:57

Edited BY: Meni Shay Edit Date: 2017-05-22 17:34:56

Edited BY: Meni Shay Edit Date: 2017-05-22 17:32:14

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