A body of weight 4N rest on a plane inclined at 30degree to the horizontal. What force keep it from sliding from the plane?

F = Mg*sin A = 4*sin30 = ___ N.

To determine the force that keeps the body from sliding down the inclined plane, we need to consider the forces acting on it.

1. Weight: The weight of the body acts vertically downward and is given as 4N. We can split this weight into two components: one parallel to the plane (down the plane) and one perpendicular to the plane.

Weight parallel to the plane = weight * sin(angle of inclination)
Weight parallel to the plane = 4N * sin(30°) = 2N

Weight perpendicular to the plane = weight * cos(angle of inclination)
Weight perpendicular to the plane = 4N * cos(30°) ≈ 3.46N (rounded to two decimal places)

2. Normal Force: The normal force acts perpendicular to the plane, and it counterbalances the perpendicular component of the weight. It can be calculated as the weight component perpendicular to the plane, which is 3.46N in this case.

3. Friction Force: The force that keeps the body from sliding down the inclined plane is the static friction force. It acts parallel to the plane and opposes the component of the weight down the plane.

Since the body is at rest, the static friction force must be equal to the weight component down the plane. Therefore, the force that keeps the body from sliding is:

Force keeping the body from sliding = Weight parallel to the plane = 2N

So, a force of 2N is required to keep the 4N body from sliding down the inclined plane.

To find the force that keeps the body from sliding down the inclined plane, we need to calculate the component of the weight that acts parallel to the plane.

First, let's draw a diagram to visualize the problem:

```
|
|
/|
/ |
/ | <- inclined plane
/ |
/ |
/ |
/ |
/body /|
```

We can break the weight (W) of the body into two components:

1. The component perpendicular to the plane (W ⊥)
2. The component parallel to the plane (W ||)

In this case, the weight of the body (W) is equal to 4N.

The component perpendicular to the plane (W ⊥) can be found using the formula:

W ⊥ = W * cos(θ)

Where θ is the angle of inclination, which is 30 degrees in this case.

W ⊥ = 4N * cos(30°)

W ⊥ = 4N * 0.866

W ⊥ ≈ 3.464N

Now, to find the force that keeps the body from sliding down the plane, we need to consider the component parallel to the plane (W ||). This force, also known as the force of friction, counteracts the component of the weight trying to make the body slide.

The force of friction (F) can be found using the formula:

F = μ * N

Where μ is the coefficient of friction and N is the normal force.

In this case, the normal force (N) is equal to the component perpendicular to the plane (W ⊥):

N = W ⊥

N ≈ 3.464N

Finally, we can substitute the values into the formula to find the force of friction:

F = μ * N

F = μ * 3.464N

Since the problem doesn't mention the coefficient of friction, we can't determine the exact force of friction without additional information. The coefficient of friction depends on the nature of the surfaces in contact, and different materials will have different coefficients of friction.

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