Stuck on ladder problem, know the solution but need an explanation

[BifSlamkovich]

Homework Statement

http://www.webassign.net/walker/p11-78alt.gif
An 85 kg person stands on a uniform ladder that is 4.3 m long and weighs 65.0 N. The floor is rough; hence it exerts both a normal force, f1, and a frictional force, f2, on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f3. Use the dimensions in the figure to find the following. (a = 3.8 m.)
f1=?
f2=?
f3=?

Homework Equations

I know the solution but I don't understand why f₃(3.8)-mg(2.15)=0. How is this a correct expression of the torque?

The Attempt at a Solution

 

[PhanthomJay]

Homework Statement

http://www.webassign.net/walker/p11-78alt.gif
An 85 kg person stands on a uniform ladder that is 4.3 m long and weighs 65.0 N. The floor is rough; hence it exerts both a normal force, f1, and a frictional force, f2, on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f3. Use the dimensions in the figure to find the following. (a = 3.8 m.)
f1=?
f2=?
f3=?

Homework Equations

I know the solution but I don't understand why f₃(3.8)-mg(2.15)=0. How is this a correct expression of the torque?

The Attempt at a Solution

It is not. The correct expression for torque about the base of the ladder must include f3, the man's weight, and the ladder weight, using the appropriate perpendicular diatances from those forces to the ladder base to calculate the respective torques and signs of the torques. You are missing a few dimensions.

 

[thomate1]

First of all, it is important to understand that torque is the measure of the force that can cause an object to rotate around an axis or pivot point. In this case, the axis of rotation is at the point where the ladder touches the ground, and the pivot point is where the ladder touches the wall.

To find the solution for f3, we need to use the equation for torque, which is given by τ = r x F, where τ is the torque, r is the distance from the pivot point, and F is the force applied.

In this problem, the torque due to f3 is equal to the weight of the ladder (mg) multiplied by the distance between the pivot point and the point where f3 is applied (which is 3.8 m). This is because the force of gravity is acting downwards on the ladder, and it is balanced by the normal force (f3) acting upwards at the pivot point. The equation can be written as τ = f3(3.8) - mg(2.15) = 0, since the ladder is not rotating, the net torque must be equal to zero.

Similarly, for f1, the torque is given by τ = f1(4.3) - mg(2.15) = 0. This is because f1 is acting at a distance of 4.3 m from the pivot point, and it is also balancing the weight of the ladder.

For f2, we need to consider both the normal force and the frictional force. The normal force is acting at a distance of 1.9 m from the pivot point, while the frictional force is acting at a distance of 2.15 m from the pivot point. The torque equation can be written as τ = f2(2.15) + f1(1.9) - mg(2.15) = 0, since both forces are acting in opposite directions and balancing each other out.

Overall, the key to understanding the solution is to recognize that torque is a measure of the force that causes rotation, and it is dependent on the distance from the pivot point. By balancing the torques in each direction, we can find the values of f1, f2, and f3.