Where is the fulcrum located in this simple torque problem?

[Joza]

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.

 

[rl.bhat]

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

 

[ogb p]

First of all, let's define some variables to make the problem easier to understand. Let's call the force of the normal at the wall "N" and the distance from the base of the ladder to the center of mass "d". We can also define the distance from the wall to the center of mass as "x".

Now, let's look at the forces acting on the ladder. We have the weight of the ladder, which is equal to Mg, acting downwards. We also have the normal force N acting perpendicular to the wall, which is what keeps the ladder in equilibrium.

The equation given for the sum of torques is correct. The first term, 3Mg, represents the torque due to the weight of the ladder, which is acting at a distance of 3m from the base (since the base is 6m and the center of mass is at 3m). The second term, F8, represents the torque due to the normal force, which is acting at a distance of 8m from the base (since the ladder is 10m long and the center of mass is at 8m).

The fulcrum in this problem would be at the point where the ladder makes contact with the ground, since this is the point where the ladder can rotate.

To solve this problem, we can set the sum of torques equal to zero, since the ladder is in equilibrium. This gives us the equation 3Mg - F8 = 0. We can rearrange this to solve for F, which gives us F = 3/8 Mg. This means that the normal force at the wall is 3/8 times the weight of the ladder.

I hope this explanation helps you understand the problem better. It's important to always define your variables and consider all the forces acting on an object in order to solve a physics problem accurately. Let me know if you have any further questions.