A particle of 10 kg is kept on a smooth inclined surface of an inclination 60 degrees with the horizontal. To keep the particle in equilibrium, find the horizontal force needed and... A particle of 10 kg is kept on a smooth inclined surface of an inclination 60 degrees with the horizontal. To keep the particle in equilibrium, find the horizontal force needed and find the force needed parallel to the force. Read more

Steps to Solve

1. Draw a free body diagram

Draw the inclined plane, the particle, and all the forces acting on the particle. These forces are: - Gravitational force $mg$ acting vertically downwards, where $m = 10$ kg and $g = 9.8$ m/s$^2$. - Applied horizontal force $F_H$ acting horizontally. - Normal reaction force $R$ acting perpendicular to the inclined plane. - Applied force parallel to the inclined plane $F_P$.

2. Resolve forces into components

Resolve the gravitational force $mg$ into two components: one along the inclined plane ($mg\sin\theta$) and one perpendicular to the inclined plane ($mg\cos\theta$), where $\theta = 60^\circ$. We have: - $mg\sin\theta = 10 \times 9.8 \times \sin(60^\circ)$ - $mg\cos\theta = 10 \times 9.8 \times \cos(60^\circ)$

3. Equilibrium conditions

For the particle to be in equilibrium, the net force in both the horizontal and vertical directions must be zero. Since we're looking for a horizontal force $F_H$ and a force parallel to the inclined plane $F_P$ required to keep the particle in equilibrium, we need to consider equilibrium conditions along the inclined plane. In this case, the force $F_P$ will counteract the component of gravity along the plane:

$F_P = mg\sin\theta$

4. Calculate $F_P$

Substitute the values of $m$, $g$, and $\theta$ into the equation: $F_P = 10 \times 9.8 \times \sin(60^\circ) = 10 \times 9.8 \times \frac{\sqrt{3}}{2} = 49\sqrt{3} \approx 84.87$ N

5. Determine the horizontal force $F_H$ required for equilibrium

Consider the forces perpendicular to the inclined plane. The normal reaction $R$ balances the component of gravity perpendicular to the plane plus the component of the horizontal force $F_H$ perpendicular to the plane. The forces along the inclined plane balance each other: $F_P$ balances the component of gravity along the inclined plane, and the component of $F_H$ along the inclined plane.

To find $F_H$, consider that its component along the plane must provide the horizontal equilibrium, therefore: $F_H = mg \tan\theta = 10 \times 9.8 \times \tan(60^\circ) = 10 \times 9.8 \times \sqrt{3} = 98\sqrt{3} \approx 169.74$ N