Due to friction ,the system as shown in the diagram remains motionless. Calculate the static coefficient of friction ?

enter image source here

1 Answer
Aug 16, 2017

Answer:

#mu_s<=0.346#

Explanation:

Here's what I tried.

  • I defined up the ramp as the positive direction.

Forces on #m_1#:

#sumF_x=T_1-(F_G)_x-f_s=0#

#sumF_y=n-(F_G)_y=0#

  • #f_(s"max")=mu_sn#

  • #(F_G)_x=m_1gsin(theta)#

  • #(F_G)_y=m_2gcos(theta)#

  • #n=mgcos(theta)#

I will refer to #f_(s"max")# simply as #f_s# from this point on, though I am still solving in terms of the maximum static friction.

#=>f_s=T_1-m_1gsin(theta)#

#=>mu_sm_1gcos(theta)=T_1-m_1gsin(theta)#

#=>color(darkblue)(mu_s=(T_1-m_1gsin(theta))/(m_1gcos(theta)))#

Forces on #m_2#:

#sumF=sumF_y=T_2-F_G=0#

  • #F_G=m_2g#

Because we can assume a massless rope and frictionless pulley, #vecT_1# and #vecT_2# act as an "action/reaction" pair.

  • #T_1=T_2=m_2g#

#=>mu_s=(m_2g-m_1gsin(theta))/(m_1gcos(theta))#

#=>mu_s=(cancel(g)(m_2-m_1sin(theta)))/(cancel(g)(m_1cos(theta))#

#=>color(darkblue)(mu_s=(m_2-m_1sin(theta))/(m_1cos(theta)))#

Using known values:

#mu_s=(80-100(0.500))/(100(0.866))#

#mu_(s"max")=0.346#

#=>mu_s<=0.346#