Question

Find acceleration of m when system accelerates.

M

m

Smooth

Answer 1

\frac{mg\sqrt{2}}{M + 2m}

Answer 2

Explanation of the solution:

1. Define Accelerations: Let $A$ be the horizontal acceleration of block $M$. Block $m$ moves along the vertical face of $M$, so its horizontal acceleration is also $A$. Let $a_{rel}$ be the downward acceleration of $m$ relative to $M$. The absolute vertical acceleration of $m$ is $a_{rel}$.
2. String Constraint: The string connects a fixed point on the ground to the pulley on $M$, and then to $m$. The total length of the string is constant. Let $x_M$ be the position of $M$ and $y_m$ be the vertical position of $m$. Differentiating the string length equation, $L = (x_M - x_{fixed}) + (H - y_m)$, twice with respect to time yields $0 = a_M - a_y$. Thus, the horizontal acceleration of $M$ ($A$) is equal to the absolute downward vertical acceleration of $m$ ($a_{rel}$). So, $A = a_{rel}$.
3. Free Body Diagrams and Newton's Second Law:
   • For block $m$:
     • Vertical forces: $mg$ (downwards), $T$ (upwards). Equation: $mg - T = ma_{rel} = mA$.
     • Horizontal forces: $N_M$ (normal force from $M$ on $m$, to the right). Equation: $N_M = mA$.
   • For block $M$:
     • Horizontal forces: $T$ (from horizontal string, to the right), $N_m$ (normal force from $m$ on $M$, to the left). By Newton's third law, $N_m = N_M$. Equation: $T - N_M = MA$.
4. Solve the System of Equations:
   • From $N_M = mA$, substitute into the equation for $M$: $T - mA = MA \implies T = (M+m)A$.
   • Substitute $T$ into the equation for $m$: $mg - (M+m)A = mA$.
   • Rearrange and solve for $A$: $mg = mA + MA + mA = (M+2m)A \implies A = \frac{mg}{M+2m}$.
5. Acceleration of $m$: Block $m$ has a horizontal acceleration $A$ and a vertical acceleration $A$. The magnitude of its total acceleration is $a_m = \sqrt{A^2 + A^2} = A\sqrt{2}$.
6. Final Result: Substitute the value of $A$: $a_m = \frac{mg\sqrt{2}}{M+2m}$.

The final answer is $\boxed{\frac{mg\sqrt{2}}{M + 2m}}$.