A block of mass m is placed on a smooth wedge of incination `theta`. The whole system s acelerated horizontally so tht the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude

Options

A. mg
B. `mg/costheta`
C. `mg costheta`
D. `mg tantheta`

Answer

To solve the problem of finding the force exerted by the wedge on the block, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The block of mass \( m \) experiences the following forces:
- **Gravitational Force (Weight)**: \( \vec{W} = mg \) acting downward.
- **Normal Force (N)**: Exerted by the wedge, acting perpendicular to the surface of the wedge.
- **Pseudoforce (F_p)**: Since the system is accelerating horizontally, a pseudoforce acts on the block in the opposite direction of the acceleration \( a \).

### Step 2: Analyze the Forces in the Inclined Plane Coordinates
We will resolve the forces into components parallel and perpendicular to the inclined plane:
- The gravitational force can be resolved into:
- Perpendicular to the incline: \( W_{\perp} = mg \cos \theta \)
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)

- The pseudoforce can also be resolved:
- Perpendicular to the incline: \( F_{p\perp} = ma \sin \theta \)
- Parallel to the incline: \( F_{p\parallel} = ma \cos \theta \)

### Step 3: Apply the Equilibrium Conditions
Since the block does not slip, it is in equilibrium in both the directions parallel and perpendicular to the incline.

#### For the parallel direction:
\[ \Sigma F_{\parallel} = 0 \implies mg \sin \theta - ma \cos \theta = 0 \]
From this, we can derive:
\[ mg \sin \theta = ma \cos \theta \implies a = g \frac{\sin \theta}{\cos \theta} = g \tan \theta \]

#### For the perpendicular direction:
\[ \Sigma F_{\perpendicular} = 0 \implies N - mg \cos \theta - ma \sin \theta = 0 \]
Substituting \( a \) from the previous result:
\[ N = mg \cos \theta + m(g \tan \theta) \sin \theta \]
This simplifies to:
\[ N = mg \cos \theta + mg \sin^2 \theta \]

### Step 4: Simplify the Expression for Normal Force
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ N = mg (\cos^2 \theta + \sin^2 \theta) = mg \]
Thus, we can express \( N \) as:
\[ N = \frac{mg}{\cos \theta} \]

### Final Answer
The force exerted by the wedge on the block is:
\[ N = \frac{mg}{\cos \theta} \]