Question

A body of mass \(2 \, {kg}\) is placed on a smooth horizontal surface. Two forces \(F_1 = 20 \, {N}\) and \(F_2 = 10\sqrt{3} \, {N}\) are acting on the body in the directions making angles of \(30^\circ\) and \(60^\circ\) to the surface. The reaction of the surface on the body is: 

 

• A. \(20 \, {N}\)
• B. \(25 \, {N}\)
• C. \(5 \, {N}\)
• D. Zero

Hint:

In problems involving forces on horizontal surfaces, always check for the vertical components of all forces to determine the correct normal reaction.

Answer 1

The Correct Option is D

The forces \( F_1 \) and \( F_2 \) are acting at angles to the horizontal surface. We need to find the reaction force exerted by the surface.

The horizontal components of the forces \( F_1 \) and \( F_2 \) do not affect the vertical reaction of the surface because they do not contribute to the vertical force.

The vertical components of the forces \( F_1 \) and \( F_2 \) are responsible for the reaction from the surface. The vertical component of each force is given by:

\[ F_{1v} = F_1 \sin 30^\circ = 20 \times \frac{1}{2} = 10 \, \text{N} \] \[ F_{2v} = F_2 \sin 60^\circ = 10\sqrt{3} \times \frac{\sqrt{3}}{2} = 15 \, \text{N} \]

The total vertical force acting on the body is the sum of the vertical components of the forces \( F_1 \) and \( F_2 \):

\[ F_{\text{total vertical}} = 10 + 15 = 25 \, \text{N} \]

The body is in equilibrium, so the reaction force from the surface must balance the total vertical force, which would typically be \( 25 \, \text{N} \) downward.

However, because the surface is smooth and the body is not moving vertically, the net reaction from the surface in the vertical direction is zero due to the exact cancellation of forces acting on the body.

Thus, the reaction force is zero.

Answer 2

To solve the problem, we need to calculate the reaction force exerted by the surface on the body, considering the two applied forces at angles to the surface.

1. Understanding the Forces:
The forces \( F_1 \) and \( F_2 \) act on the body at angles of 30° and 60° to the horizontal surface. The magnitudes of the forces are given as:

\( F_1 = 20 \, \text{N} \)

\( F_2 = 10 \sqrt{3} \, \text{N} \)

2. Resolving the Forces into Components:
Since the forces are acting at angles, we need to resolve them into their horizontal and vertical components:
For \( F_1 \):

Horizontal component: \( F_{1x} = F_1 \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 20 \sqrt{3}/2 \)

Vertical component: \( F_{1y} = F_1 \sin 30^\circ = 20 \times \frac{1}{2} = 10 \) - For \( F_2 \):

Horizontal component: \( F_{2x} = F_2 \cos 60^\circ = 10 \sqrt{3} \times \frac{1}{2} = 5 \sqrt{3} \)

Vertical component: \( F_{2y} = F_2 \sin 60^\circ = 10 \sqrt{3} \times \frac{\sqrt{3}}{2} = 15 \)

3. Considering the Smooth Surface:
The problem states that the body is placed on a smooth horizontal surface. Therefore, there is no friction. The vertical and horizontal components of the forces acting on the body should balance out in such a way that the reaction force exerted by the surface is zero.

4. Net Force in the Vertical Direction:
Since the body is on a smooth surface and there is no friction, the surface does not exert any force in response to the applied forces. In fact, the net vertical and horizontal forces would balance each other out, leading to no net reaction force on the body.

Final Answer:
The correct answer is Option 4: Zero, as the body experiences no reaction force on the smooth surface under the given conditions.