A block of mass 5 kg is placed on a rough inclined surface as shown in the figure.

If $\vec{F_1}$ is the force required to just move the block up the inclined plane and $\vec{F_2}$ is the force required to just prevent the block from sliding down, then the value of $|\vec{F_1}| - |\vec{F_2}|$ is: [Use $g = 10 \, \text{m/s}^2$]

Question

A solid sphere of mass $5$ kg and radius $10$ cm is kept in contact with another solid sphere of mass $10$ kg and radius $20$ cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ___________ kg$\cdot$m$^2$.

Answer 1

To resolve this issue, we must compute the magnitudes of forces $|\vec{F_1}|$ and $|\vec{F_2}|$ and subsequently determine their difference, $|\vec{F_1}| - |\vec{F_2}|$.

Provided data:

Block mass ($m$) = 5 kg
Inclination angle ($\theta$) = $30^\circ$
Friction coefficient ($\mu$) = 0.1
Gravitational acceleration ($g$) = 10 m/s²

The forces acting on the block, parallel and perpendicular to the inclined plane, are:

$F_{\text{gravity parallel}} = mg \sin(\theta)$

$F_{\text{gravity perpendicular}} = mg \cos(\theta)$

The maximum static friction is calculated as:

$F_{\text{friction}} = \mu \cdot F_{\text{gravity perpendicular}} = \mu \cdot mg \cos(\theta)$

Calculation of $|\vec{F_1}|$ (force required to move the block upwards):

The force to move the block upwards must counteract both gravitational pull and friction:

$|\vec{F_1}| = F_{\text{gravity parallel}} + F_{\text{friction}}$

$= mg \sin(\theta) + \mu mg \cos(\theta)$

Substitution of values yields:

$|\vec{F_1}| = 5 \cdot 10 \cdot \sin(30^\circ) + 0.1 \cdot 5 \cdot 10 \cdot \cos(30^\circ)$

$= 50 \cdot 0.5 + 0.1 \cdot 50 \cdot \frac{\sqrt{3}}{2}$

$= 25 + 2.5\sqrt{3}$

Calculation of $|\vec{F_2}|$ (force to prevent the block from sliding downwards):

To prevent downward motion, the applied force must balance the component of gravity acting down the incline, opposed by friction:

$|\vec{F_2}| = F_{\text{gravity parallel}} - F_{\text{friction}}$

$= mg \sin(\theta) - \mu mg \cos(\theta)$

Substituting the given values:

$|\vec{F_2}| = 5 \cdot 10 \cdot \sin(30^\circ) - 0.1 \cdot 5 \cdot 10 \cdot \cos(30^\circ)$

$= 25 - 2.5\sqrt{3}$

Final Difference Calculation:

$|\vec{F_1}| - |\vec{F_2}| = (25 + 2.5\sqrt{3}) - (25 - 2.5\sqrt{3})$

$= 5\sqrt{3} \, \text{N}$

Therefore, the final result is $5\sqrt{3} \, \text{N}$.

The Correct Option is B

Solution and Explanation

Calculation of \(|\vec{F_1}|\) (force required to move the block upwards):

Calculation of \(|\vec{F_2}|\) (force to prevent the block from sliding downwards):

Final Difference Calculation:

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