Static and kinetic friction
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?

3 People tried to answer this question

mu_s>0.58 because in the case described here it must be that tan(alpha)<mu_s

mu_s>0.5 since static friction coefficient is always larger than the kinetic friction coefficient.

It is always the case that 0<mu_s<1

The scenario is not realistic: What comes up must come down. No limit can be placed on mu_s

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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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