To solve the problem, we need to analyze the forces acting on the wedge and the blocks placed on it. Here's a step-by-step solution:
### Step 1: Identify the masses and angles
- Mass of each block, \( m = 2 \, \text{kg} \)
- Mass of the wedge, \( M = 12 \, \text{kg} \)
- Coefficient of friction between the wedge and the ground, \( \mu = 0.3 \)
- The angle of the wedge is \( \theta = 30^\circ \) (since the blocks are resting on the wedge which is inclined).
### Step 2: Calculate the gravitational force acting on the blocks
The gravitational force acting on each block is given by:
\[
F_g = m \cdot g
\]
where \( g \approx 9.81 \, \text{m/s}^2 \).
For each block:
\[
F_{gA} = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19.62 \, \text{N}
\]
\[
F_{gB} = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19.62 \, \text{N}
\]
### Step 3: Determine the normal forces acting on the blocks
The normal force \( N \) acting on each block can be calculated using the angle of the wedge. The normal force for each block is given by:
\[
N_A = m \cdot g \cdot \cos(30^\circ)
\]
\[
N_B = m \cdot g \cdot \cos(30^\circ)
\]
Calculating \( N_A \) and \( N_B \):
\[
N_A = 2 \cdot 9.81 \cdot \cos(30^\circ) \approx 2 \cdot 9.81 \cdot 0.866 \approx 17.02 \, \text{N}
\]
\[
N_B = 2 \cdot 9.81 \cdot \cos(30^\circ) \approx 17.02 \, \text{N}
\]
### Step 4: Calculate the horizontal components of the normal forces
The horizontal components of the normal forces acting on the wedge due to the blocks are:
\[
F_{horizontalA} = N_A \cdot \sin(30^\circ) = 17.02 \cdot 0.5 = 8.51 \, \text{N}
\]
\[
F_{horizontalB} = N_B \cdot \sin(30^\circ) = 17.02 \cdot 0.5 = 8.51 \, \text{N}
\]
### Step 5: Determine the net horizontal force on the wedge
Since both horizontal components act in opposite directions, they cancel each other out:
\[
F_{net} = F_{horizontalA} - F_{horizontalB} = 8.51 - 8.51 = 0 \, \text{N}
\]
### Step 6: Calculate the maximum static friction force between the wedge and the ground
The maximum static friction force \( F_{friction} \) that can act on the wedge is given by:
\[
F_{friction} = \mu \cdot N_{wedge}
\]
where \( N_{wedge} \) is the normal force acting on the wedge due to its weight:
\[
N_{wedge} = M \cdot g = 12 \cdot 9.81 \approx 117.72 \, \text{N}
\]
Thus,
\[
F_{friction} = 0.3 \cdot 117.72 \approx 35.32 \, \text{N}
\]
### Step 7: Conclusion
Since the net horizontal force on the wedge is zero, the frictional force exerted by the ground on the wedge will also be zero because there is no net force trying to move the wedge.
### Final Answer
The frictional force by the ground on the wedge is:
\[
\text{Frictional Force} = 0 \, \text{N}
\]
