These expressions are combined (to eliminate F') and we arrive at FmgmmMs=−+FHGIKJμ1which we find to be F= 4.9 ×102N. 33. The free-body diagrams for the two blocks, treated individually, are shown below (firstmand then M).F'is the contact force between the two blocks, and the static friction forceGfsis at its maximum value (so Eq. 6-1 leads to fs= fs,max= sF'wheres= 0.38). Treating the two blocks together as a single system (sliding across a frictionless floor), we apply Newton’s second law (with +
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