Question

A body of mass \(10\,\text{kg}\) is moving up on an inclined plane of \(30^\circ\) with an acceleration \(2\,\text{m s}^{-2}\). Find the force required, if the coefficient of friction between the object and the plane surface is \(\frac{\sqrt{3}}{6}\), [given \(g = 10\,\text{m s}^{-2}\)].

• A. \(1.95\,\text{N}\)
• B. \(9.5\,\text{N}\)
• C. \(19\,\text{N}\)
• D. \(95\,\text{N}\)

Hint:

When a body moves up an inclined plane, both friction and the component \(mg\sin\theta\) act downward along the plane.

Answer 1

The Correct Option is D

Step 1: Identify the forces acting on the body.
The body is moving up the inclined plane, so friction acts downward along the plane. The component of weight \(mg\sin\theta\) also acts downward along the plane.

Step 2: Write the equation of motion along the inclined plane.
\[ F - mg\sin\theta - f = ma \]
where friction force is:
\[ f = \mu mg\cos\theta \]

Step 3: Substitute friction in the equation.
\[ F = ma + mg\sin\theta + \mu mg\cos\theta \]

Step 4: Substitute given values.
\[ m = 10\,\text{kg},\quad a = 2\,\text{m s}^{-2},\quad g = 10\,\text{m s}^{-2} \]
\[ \theta = 30^\circ,\quad \mu = \frac{\sqrt{3}}{6} \]
\[ F = 10 \times 2 + 10 \times 10 \times \sin 30^\circ + \frac{\sqrt{3}}{6} \times 10 \times 10 \times \cos 30^\circ \]

Step 5: Use trigonometric values.
\[ \sin 30^\circ = \frac{1}{2} \]
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]
\[ F = 20 + 100 \times \frac{1}{2} + \frac{\sqrt{3}}{6} \times 100 \times \frac{\sqrt{3}}{2} \]

Step 6: Simplify the terms.
\[ F = 20 + 50 + \frac{300}{12} \]
\[ F = 20 + 50 + 25 \]
\[ F = 95\,\text{N} \]

Step 7: Final answer.
\[ \boxed{95\,\text{N}} \]