Where is the fulcrum located in this simple torque problem?

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In the torque problem involving a 10m ladder leaning against a frictionless wall, the forces acting on the ladder include its weight, the normal force from the wall, and the ground reaction force. The ladder's weight acts downward through its center of mass, while the normal force at the wall acts horizontally. The equation for the sum of torques is given as 3Mg - F8, indicating the moments created by these forces. To find the fulcrum's location, one must analyze the balance of clockwise and anticlockwise moments around a pivot point. Understanding these forces and their moments is essential for solving the equilibrium condition of the ladder.
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This is a pretty simple problem, I am just a bit confused.

Consider a 10m-long homogeneous ladder of mass M that leans in equilibrium against a vertical frictionless wall. Identify the forces acting on the ladder and evaluate their magnitude (relative to the weight W=Mg).

I can't draw the diagram.

However, the length of the base subtended by the ladder to the wall is 6m, and so the height is 8m.

The equation given for sum of torques is 3Mg - F8, where F is the force of the normal at the wall. Where abouts would the fulcrum be? Maybe then i can understand this equation.
 
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since the ladder is in equilibrium, a force equal to F must be acting parallel to the ground at the bottom of the ladder towards the wall. Wieght of the ladder acts through center of mass and vertically downward. A normal reaction equal to weight of the ladder acts at the end of the ladder in the vertical direction. Now consider clockwise moment and anticlockwise moment and find the sum
 
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