A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination 𝜃 = 30° from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance 𝑟 = 0.5 m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ____ 𝑚𝑠��. (Take 𝑔 = 10 𝑚 𝑠��.)

 We are given a solid sphere of mass 

$m=1\text{kg}$ and radius $R=1\text{m}$ rolling without slipping on a fixed inclined plane with an angle of inclination $\theta =30^{\circ }$. Two forces of magnitude $F=1\text{N}$ each, parallel to the incline, act on the sphere at a distance $r=0.5\text{m}$ from the center of the sphere. We are to find the acceleration of the sphere down the plane.

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Step 1: Free-body diagram and forces

The forces acting on the sphere are:

1. Gravitational force: $mg\sin \theta$ acting down the incline.

2. Normal force: $N=mg\cos \theta$ acting perpendicular to the incline.

3. Frictional force: $f$ acting up the incline (providing the torque for rolling without slipping).

4. Two applied forces: Each of magnitude $F=1\text{N}$, acting parallel to the incline at a distance $r=0.5\text{m}$ from the center.

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Step 2: Net force along the incline

The net force along the incline is:

$$F_{\text{net}}=mg\sin \theta +2F-f\mathrm{.}$$

Substitute $m=1\text{kg}$, $g=10{\text{m/s}}^2$, $\theta =30^{\circ }$, and $F=1\text{N}$:

$$F_{\text{net}}=(1)(10)(\sin 30^{\circ })+2(1)-f=5+2-f=7-f\mathrm{.}$$

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Step 3: Torque and rotational motion

The sphere rolls without slipping, so the linear acceleration $a$ and angular acceleration $\alpha$ are related by:

$$a=R\alpha \mathrm{.}$$

The torque $\tau$ about the center of the sphere is due to the frictional force $f$ and the two applied forces $F$:

$$\tau =fR-2Fr\mathrm{.}$$

For a solid sphere, the moment of inertia $I$ is:

$$I=\frac{2}{5}m{R}^2\mathrm{.}$$

Using $\tau =I\alpha$, we get:

$$fR-2Fr=\left(\frac{2}{5}m{R}^2\right)\alpha \mathrm{.}$$

Substitute $\alpha =\frac{a}{R}$, $m=1\text{kg}$, $R=1\text{m}$, $F=1\text{N}$, and $r=0.5\text{m}$:

$$f(1)-2(1)(0.5)=(\frac{2}{5}(1)(1{)}^2)\left(\frac{a}{1}\right),$$$$f-1=\frac{2}{5}a\mathrm{.}$$

Solve for $f$:

$$f=\frac{2}{5}a+1.$$

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Step 4: Solve for acceleration

From Step 2, the net force is:

$$F_{\text{net}}=7-f\mathrm{.}$$

Substitute $f=\frac{2}{5}a+1$:

$$F_{\text{net}}=7-(\frac{2}{5}a+1)=6-\frac{2}{5}a\mathrm{.}$$

Using Newton's second law $F_{\text{net}}=ma$:

$$6-\frac{2}{5}a=(1)a,$$$$6=a+\frac{2}{5}a,$$$$6=\frac{7}{5}a\mathrm{.}$$

Solve for $a$:

$$a=\frac{6\cdot 5}{7}=\frac{30}{7}{\text{m/s}}^2\mathrm{.}$$

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Final Answer:

The acceleration of the sphere down the plane is:

$$a=\frac{30}{7}{\text{m/s}}^2\mathrm{.}$$