A wedge of mass M = 3.0 kg rests on a smooth horizontal floor…

Step 1. Forces on the block, in the wedge's frame. The wedge accelerates towards the slope side with acceleration $A$. In its frame the block feels a pseudo force $mA$ the other way. This force has a part $mA\cos\theta$ up the slope and a part $mA\sin\theta$ into the slope. Gravity gives $mg\sin\theta$ down the slope and $mg\cos\theta$ into it. So the normal force is

Step 2. The lower limit. For a small $A$ the block tends to slide down, so friction acts up the slope, at most $\mu N$. On the point of slipping,

$mg\sin\theta = mA\cos\theta + \mu m(g\cos\theta + A\sin\theta)$, so

While the block stays on the wedge, the two move as one body: $F = (M + m)A = 4.0\ \text{kg} \times 4.21\ \text{m/s}^2 = 16.8\ \text{N}$.

Step 3. The upper limit. Now the block tends to slide up, and friction acts down the slope:

$mA\cos\theta = mg\sin\theta + \mu m(g\cos\theta + A\sin\theta)$, so

and $F = 4.0\ \text{kg} \times 12.31\ \text{m/s}^2 = 49.2\ \text{N}$. So the block stays at rest for $16.8\ \text{N} \le F \le 49.2\ \text{N}$.

Step 4. $F = 60\ \text{N}$: the block slides up. Let the wedge's acceleration be $A$, and the block's acceleration up the slope, relative to the wedge, be $b$. Sliding friction $\mu N$ acts down the slope on the block.

• Block, across the slope (it does not move across it): $N = m(g\cos\theta + A\sin\theta)$.
• Block, along the slope: $mb = mA\cos\theta - mg\sin\theta - \mu N$.
• Wedge, horizontally, in the ground frame: the block pushes back on it with $N$ and with friction, and both have horizontal parts against $F$, adding to $N(\sin\theta + \mu\cos\theta)$. So $MA = F - N(\sin\theta + \mu\cos\theta)$.

Step 5. Solve. Put $N$ into the wedge's equation:

Then, from the block's equation along the slope,

up the slope.