Two cubes of masses $m_1$ and $m_2$ be on two frictionless slopes of block $A$ which rests on a horizontal table. The cubes are connected by a string which passes over a pulley as shown in the figure. To what horizontal acceleration $f$ should the whole system (that is blocks and cubes) be subjected so that the cubes do not slide down the planes. What is the tension of the string in this situation?

Step 1: Define the forces in the accelerating frame.

Consider the non-inertial reference frame of block $A$ accelerating horizontally to the right with acceleration $f$, where a pseudo-force of magnitude $mf$ acts horizontally to the left on each mass.

Step 2: Set up the equation of motion for mass $m_2$.

Balance the forces acting on mass $m_2$ along the slope of angle $\beta$.

$$T = m_2(f\cos\beta - g\sin\beta)$$

Step 3: Set up the equation of motion for mass $m_1$.

Balance the forces acting on mass $m_1$ along the slope of angle $\alpha$.

$$T = m_1(g\sin\alpha - f\cos\alpha)$$

Step 4: Solve for the horizontal acceleration $f$.

Equate the two expressions for the tension $T$.

$$m_2(f\cos\beta - g\sin\beta) = m_1(g\sin\alpha - f\cos\alpha)$$

Rearrange the terms to group the acceleration $f$ on one side.

$$f(m_1\cos\alpha + m_2\cos\beta) = g(m_1\sin\alpha + m_2\sin\beta)$$ $$f = \frac{g(m_1\sin\alpha + m_2\sin\beta)}{m_1\cos\alpha + m_2\cos\beta}$$

Step 5: Calculate the tension $T$ in the string.

Substitute the expression for $f$ back into the equation for $T$.

$$T = m_1\left(g\sin\alpha - \left[\frac{g(m_1\sin\alpha + m_2\sin\beta)}{m_1\cos\alpha + m_2\cos\beta}\right]\cos\alpha\right)$$

Simplify the expression using the trigonometric identity $\sin\alpha\cos\beta - \cos\alpha\sin\beta = \sin(\alpha - \beta)$.

$$T = \frac{m_1 m_2 g \sin(\alpha - \beta)}{m_1\cos\alpha + m_2\cos\beta}$$

Answer: The required horizontal acceleration is $f = \dfrac{g(m_1\sin\alpha + m_2\sin\beta)}{m_1\cos\alpha + m_2\cos\beta}$ and the tension in the string is $T = \dfrac{m_1 m_2 g \sin(\alpha - \beta)}{m_1\cos\alpha + m_2\cos\beta}$.