(a) Both bodies remain motionless

For the block not to slide down the wedge, friction must balance the component of gravity along the slope.

$$ mg \sin 45^\circ \le \mu (mg \cos 45^\circ) $$ $$ \tan 45^\circ \le \mu \implies 1 \le \mu $$

Result: $\mu > 1.0$

(b) Wedge remains motionless, Block slides down

Here, the block slides ($\mu < 1$), but the wedge is held by floor friction. We analyze the forces exerted by the sliding block on the wedge to find the limit.

Forces from Block on Wedge:

• Normal Force $N’$: Pushes wedge down-right. Magnitude $N = mg \cos 45^\circ$.
• Friction Force $f’$: Block slides down the slope. Friction on block acts up the slope. By Newton’s 3rd law, Friction on the wedge acts down the slope (down-left)..

Horizontal Forces on Wedge:

$$ F_{horizontal} = N_x – f_x $$ Where $N_x = N \sin 45^\circ$ (Right) and $f_x = f \cos 45^\circ$ (Left). Since $N = mg/\sqrt{2}$ and $f = \mu N$: $$ F_{net, x} = \frac{mg}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} – \mu \frac{mg}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{mg}{2} (1 – \mu) $$

Resisting Friction from Floor:

Max friction on floor $f_{floor} = \mu N_{floor}$. $$ N_{floor} = Mg + N_y + f_y = mg + \frac{mg}{2} + \frac{\mu mg}{2} = mg \left( \frac{3+\mu}{2} \right) $$

Condition for Stability: Driving Force $\le$ Resisting Friction

$$ \frac{mg}{2}(1-\mu) \le \mu \cdot mg \left( \frac{3+\mu}{2} \right) $$ $$ 1 – \mu \le 3\mu + \mu^2 $$ $$ \mu^2 + 4\mu – 1 \ge 0 $$ Roots are $\mu = -2 \pm \sqrt{5}$. Positive root is $\sqrt{5} – 2$. So $\mu \ge \sqrt{5} – 2$.

Result: $\sqrt{5} – 2 < \mu < 1.0$

(c) Block slides down, Wedge slides on floor

This happens when the floor friction is insufficient to hold the wedge.

From the inequality above, this occurs when $\mu^2 + 4\mu – 1 < 0$.

Result: $\mu < \sqrt{5} - 2$

(d) Block does not slide down, but Wedge slides on floor

This is not possible