A 5.00-kg block is placed on top of a 10.0-kg block (Fig. P5.68). A horizontal force of 45.0 N is applied

to the 10.0-kg block, and the 5.00-kg block is tied to the wall. The coefficient of kinetic friction between

all surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action–reaction

forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration

of the 10.0-kg block.

In[2886]:= sumForcesX@forces_D :=

Apply@Plus, Map@Function@elt, elt@@1DD * Cos@elt@@2DDDD, forcesDDsumForcesY@forces_D :=

Apply@Plus, Map@Function@elt, elt@@1DD * Sin@elt@@2DDDD, forcesDD

In[2888]:= forces1 = 88F, 0 °<, 8Fn1, 90 °<, 8Ff1, 180 °<, 8Ff2, 180 °<, 8Fg1, 270 °<, 8Fg2, 270 °<<;

forces2 = 88Ff2, 0 °<, 8Fn2, 90 °<, 8T, 180 °<, 8Fg2, 270 °<<;

In[2948]:= 8sumForcesX@forces1D � m1 * a,

sumForcesY@forces1D � m1 * 0,

sumForcesX@forces2D � m2 * 0,

sumForcesY@forces2D � m2 * 0,

Ff1 � Fn1 * Μk,

Ff2 � Fn2 * Μk,

Fg1 � m1 * g,

Fg2 � m2 * g,

m1 ¹ 0, m2 ¹ 0, Μk ¹ 0

<;

Reduce@%, TD% �. 8m1 ® 10, m2 ® 5, Μk ® 0.2, g ® 9.8, F ® 45<

Out[2949]= Fn2 � g m2 && Fn1 � g Hm1 + m2L && Fg2 � g m2 && Fg1 � g m1 && Ff2 � g m2 Μk &&

Ff1 � g Hm1 + m2L Μk && m1 ¹ 0 && a �

F - g m1 Μk - 2 g m2 Μk

m1

&& T � g m2 Μk && m2 Μk ¹ 0

Out[2950]= Fn2 � 49. && Fn1 � 147. && Fg2 � 49. &&

Fg1 � 98. && Ff2 � 9.8 && Ff1 � 29.4 && a � 0.58 && T � 9.8