By Newton's second law, assuming <em>F</em> is horizontal,
• the net <u>horizontal</u> force on the <u>larger</u> block is
<em>F</em> - <em>µmg</em> = 3<em>mA</em>
where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;
• the net <u>vertical</u> force on the <u>larger</u> block is
4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0
where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;
• the net <u>horizontal</u> force on the <u>smaller</u> block is
<em>µmg</em> = <em>ma</em>
where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;
• the net <u>vertical</u> force on the <u>smaller</u> block is
<em>mg</em> - <em>mg</em> = 0
where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.
(You should be able to draw your own FBD's based on the forces mentioned above.)
(c) Solve the equations above for <em>A</em> and <em>a</em> :
<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)
<em>a</em> = <em>µg</em>
