A solid sphere of mass $1\,\text{kg}$ and radius $1\,\text{m}$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^\circ$ from the horizontal. Two forces of magnitude $1\,\text{N}$ each, parallel to the incline, act on the sphere, both at distance $r = 0.5\,\text{m}$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ___$\text{m s}^{-2}$. (Take $g = 10\,\text{m s}^{-2}$.)

Step 1: Identify the given parameters.

The physical properties of the solid sphere and the inclined plane are given as follows.

$$m = 1\,\text{kg}$$ $$R = 1\,\text{m}$$ $$\theta = 30^\circ$$ $$g = 10\,\text{m s}^{-2}$$ $$r = 0.5\,\text{m}$$ $$F = 1\,\text{N}$$

Step 2: Analyze the forces and the torque acting on the sphere.

The two applied forces of magnitude $F = 1\,\text{N}$ are parallel to the incline but act in opposite directions, forming a couple.

$$F_{\text{net}} = 0$$

The torque produced by this couple about the center of the sphere opposes the downward rolling motion.

$$\tau_{\text{couple}} = 2rF$$ $$\tau_{\text{couple}} = 2(0.5)(1) = 1\,\text{N m}$$

Step 3: Calculate the moment of inertia about the contact point.

Using the parallel axis theorem, the moment of inertia $I_O$ about the instantaneous point of contact $O$ is determined.

$$I_O = I_{\text{cm}} + mR^2$$ $$I_O = \frac{2}{5}mR^2 + mR^2$$ $$I_O = \frac{7}{5}mR^2$$

Step 4: Set up the torque equation about the contact point.

The net torque about the contact point is due to the component of gravity along the incline and the opposing applied couple.

$$\tau_{\text{net}} = mgR\sin\theta - 2rF$$

Apply Newton's second law for rotation about the contact point.

$$I_O \alpha = mgR\sin\theta - 2rF$$

Step 5: Relate angular acceleration to linear acceleration.

For rolling without slipping, the linear acceleration $a$ down the incline is related to the angular acceleration $\alpha$.

$$a = \alpha R$$

Substitute $\alpha = \frac{a}{R}$ and $I_O = \frac{7}{5}mR^2$ into the torque equation.

$$\frac{7}{5}mR^2 \left(\frac{a}{R}\right) = mgR\sin\theta - 2rF$$ $$\frac{7}{5}mRa = mgR\sin\theta - 2rF$$

Step 6: Solve for the acceleration $a$.

Divide both sides of the equation by $mR$.

$$\frac{7}{5}a = g\sin\theta - \frac{2rF}{mR}$$ $$a = \frac{5}{7}\left(g\sin\theta - \frac{2rF}{mR}\right)$$

Step 7: Substitute the given values.

Evaluate the terms inside the parentheses.

$$g\sin\theta = 10 \sin 30^\circ = 5$$ $$\frac{2rF}{mR} = \frac{2(0.5)(1)}{(1)(1)} = 1$$

Substitute these values to find the acceleration.

$$a = \frac{5}{7}(5 - 1)$$ $$a = \frac{20}{7}\,\text{m s}^{-2}$$

Answer: The acceleration of the sphere down the plane is $\dfrac{20}{7}\,\text{m s}^{-2}$ (or approximately $2.86\,\text{m s}^{-2}$).