On a wedge of mass 2m, a block of mass m is sliding as shown in the figure. There is no friction between block and wedge. Then the minimum coefficient of friction between wedge and ground so that the wedge does not move is:
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Resolving forces along the inclined plane is highly efficient. The horizontal component of the normal reaction force is what tries to push the wedge sideways, while its vertical component increases the effective weight of the wedge.
Step 1: Understand the setup. A block of mass $m$ slides down a smooth $45^{\circ}$ face of a wedge of mass $2m$. The face is frictionless, but there is friction between the wedge and the floor. We want the smallest floor friction that keeps the wedge still. Step 2: Find the push of the block on the wedge. The block presses on the slanted face with a normal force. For a smooth incline at angle $45^{\circ}$: \[ N_1 = mg\cos 45^{\circ} = \frac{mg}{\sqrt{2}} \] Step 3: Break this push into sideways and downward parts. This force is along the slope normal, so split it. Both parts come out equal: \[ F_h = N_1\sin 45^{\circ} = \frac{mg}{2}, \qquad F_v = N_1\cos 45^{\circ} = \frac{mg}{2} \] The sideways part $F_h$ tries to slide the wedge. The downward part $F_v$ presses the wedge onto the floor. Step 4: Find the total push on the floor. The floor must hold up the wedge weight $2mg$ plus the extra downward part from the block: \[ N_g = 2mg + \frac{mg}{2} = \frac{5mg}{2} \] Step 5: Set the no slip condition. The floor friction must be at least as big as the sideways push: \[ \mu N_g \ge F_h \;\Rightarrow\; \mu\left(\frac{5mg}{2}\right) \ge \frac{mg}{2} \] Step 6: Solve for the smallest mu. Cancel $\tfrac{mg}{2}$ from both sides: \[ 5\mu \ge 1 \;\Rightarrow\; \mu \ge \frac{1}{5} \] \[ \boxed{\mu_{\min} = 0.20} \]
