the floor of a lorry is 3m high, a plank 5m long is used as an inclined plan to raise some leading the maximum effort required to raise 200N load up the plane

two postings, still no coherent question.

To find the maximum effort required to raise a 200N load up an inclined plane, you can use the principle of equilibrium.

First, let's start by drawing a diagram to visualize the problem:

```
|\
| \
| \
| \ 5m
|________\
| 3m |\
|__________|
Floor of
Lorry
```

In this diagram, the inclined plane has a length of 5m and is placed on the floor of a lorry, which is 3m high.

Now, let's consider the forces acting on the load:

1. Weight (W = 200N): The downward force due to gravity acting on the load vertically downwards.
2. Effort (E): The force required to raise the load up the inclined plane.
3. Normal force (N): The force exerted by the inclined plane perpendicular to it.

To find the maximum effort required, we need to analyze the forces in equilibrium. In equilibrium, the sum of the forces in any direction is equal to zero.

Considering the vertical direction, we have:

Net vertical force = Weight (W) - Effort (E) * sin(θ)

Since there is no vertical acceleration (equilibrium), the net vertical force must be zero:

0 = 200N - E * sin(θ)

To determine the angle (θ), we can use trigonometry. In this case, the angle is the angle between the inclined plane and the horizontal floor. Given that the height of the lorry floor is 3m and the length of the inclined plane is 5m, we can use the trigonometric relationship:

sin(θ) = opposite/hypotenuse = 3m/5m = 0.6

Simplifying the equation, we have:

0 = 200N - E * 0.6

Rearranging the equation to solve for E, we find:

E = 200N / 0.6 = 333.33N

Therefore, the maximum effort required to raise a 200N load up the inclined plane is approximately 333.33N.