Solve this question is good English . solve this question is step and go detaily so I can understand and give reason why it is solved like that and give examples so that it can be well understood

The 4 N force acts at a distance $l$ from the pivot. The angle between this force and the plank is $30^\circ$. The component of this force perpendicular to the plank causes the moment. From the diagram, this force creates an anticlockwise moment.

$$M_{ACW} = \text{Force} \times \text{distance} \times \sin(\text{angle with plank})$$ $$M_{ACW} = 4 \text{ N} \times l \times \sin 30^\circ$$

Since $\sin 30^\circ = 0.5$:

$$M_{ACW} = 4l \times 0.5 = 2l$$

There are two forces causing clockwise moments: the weight of the plank and the 8 N force.

• Moment due to the weight of the plank: The weight of the plank is $7 \text{ N}$ and acts vertically downwards at its center of mass, which is at a distance of $\frac{L}{2}$ from the pivot (assuming a uniform plank). The perpendicular distance from the pivot to the line of action of the weight is $\frac{L}{2} \cos \theta$.

  $$M_{CW, \text{weight}} = \text{Weight} \times \text{perpendicular distance}$$ $$M_{CW, \text{weight}} = 7 \text{ N} \times \frac{L}{2} \times \cos \theta = 3.5L \cos \theta$$

• Moment due to the 8 N force: The 8 N force acts at a distance $H$ from the pivot. The angle between this force and the plank is $45^\circ$. The component of this force perpendicular to the plank causes the moment.

  $$M_{CW, 8N} = \text{Force} \times \text{distance} \times \sin(\text{angle with plank})$$ $$M_{CW, 8N} = 8 \text{ N} \times H \times \sin 45^\circ$$

  Since $\sin 45^\circ = \frac{\sqrt{2}}{2}$:

  $$M_{CW, 8N} = 8H \times \frac{\sqrt{2}}{2} = 4\sqrt{2}H$$

For the plank to be in equilibrium, the sum of anticlockwise moments must equal the sum of clockwise moments.

$$M_{ACW} = M_{CW, \text{weight}} + M_{CW, 8N}$$ $$2l = 3.5L \cos \theta + 4\sqrt{2}H$$

Rearrange the equation to solve for $\cos \theta$:

$$3.5L \cos \theta = 2l - 4\sqrt{2}H$$ $$\cos \theta = \frac{2l - 4\sqrt{2}H}{3.5L}$$

Then, $\theta$ is:

$$\theta = \arccos \left( \frac{2l - 4\sqrt{2}H}{3.5L} \right)$$

The weight acts vertically downwards. If the plank makes an angle $\theta$ with the horizontal, then the angle $\phi$ between the weight vector and the plank itself is $90^\circ - \theta$.

$$\phi = 90^\circ - \theta$$

Substituting the expression for $\theta$:

$$\phi = 90^\circ - \arccos \left( \frac{2l - 4\sqrt{2}H}{3.5L} \right)$$

Since no numerical values for $L$, $l$, and $H$ are provided in the problem, the angle must be expressed in terms of these variables.

The angle made by the weight of the plank is $\boxed{90^\circ - \arccos \left( \frac{2l - 4\sqrt{2}H}{3.5L} \right)}$.