Help to Understand Torque Equation for Ladder

[brad sue]

Hi , please help me to understand how to set up the equation for the torques.
This is the context:

A ladder of length l=5m and weigth w_ladder=160N rest against a wall. its lower end is at 3m from the wall.
A man weigthing 740N climbs 1 m along the ladder.
N_wall is the reaction ladder-wall.
(counterclock wise direction is considered positive)
The equation of the torque (equilibrium) about the bottom of the ladder is:
N_wall*4 -160* 1.5-740*1*(3/5).=0.


I don't know how they get the equation especially the underlined coefficients.
My problem is that I don't know which distance to consider when computing for the torque in this case: the distance in respect to x axis, y-axis or along the ladder.
I hope I was clear enough
Thank you
B

 

[Fermat]

Let A be the point where the ladder rests against the wall.
Let B be the point where the ladder rests against the ground.
Let C be the corner of the wall and the ground.

I assume that the wall is frictionless and that there is friction between the ladder and the ground. You probably have a coefficient of static friction, yes ?

The ladder and the wall and the ground form a triangle. This triangle is a 3-4-5 triangle. This simplifies some of the trig with cos@ = 3/5 and sin@ = 4/5, where @ is the angle between the ladder and the ground.

About your torque equation
N_wall is at A and is normal to the wall. Hence the perpindicular distance of N_wall from the ground, and hence also from the point B, is 4m, giving T₁ = N_wall*4.
The torque of w_ladder about B is the force w_ladder times the perpindicular distance from B to the line of action of w_ladder, which is 2.5m*cos@, giving T₂ = w_ladder*2.5*(3/5) = w_ladder*1.5.
The torque of the man about B is similarly given by T₃ = 740*1*cos@ = 740*1*(3/5).
Since T₂ and T₃ are clockwise torques, then they are negative.

So,

T_total = T₁ - T₂ - T₃ = N_wall*4 - w_ladder*1.5 - 740*1*(3/5).

When computing the torques, use either the x-axis or y-axis depending upon the direction of the forces.
If the force(s) is parallel to the x-axis, take the distance along the y-axis, and vice versa.
The distance should always be perpindicular to the line of action of the force.

 

[brad sue]

Let A be the point where the ladder rests against the wall.
Let B be the point where the ladder rests against the ground.
Let C be the corner of the wall and the ground.
I assume that the wall is frictionless and that there is friction between the ladder and the ground. You probably have a coefficient of static friction, yes ?
The ladder and the wall and the ground form a triangle. This triangle is a 3-4-5 triangle. This simplifies some of the trig with cos@ = 3/5 and sin@ = 4/5, where @ is the angle between the ladder and the ground.
About your torque equation
N_wall is at A and is normal to the wall. Hence the perpindicular distance of N_wall from the ground, and hence also from the point B, is 4m, giving T₁ = N_wall*4.
The torque of w_ladder about B is the force w_ladder times the perpindicular distance from B to the line of action of w_ladder, which is 2.5m*cos@, giving T₂ = w_ladder*2.5*(3/5) = w_ladder*1.5.
The torque of the man about B is similarly given by T₃ = 740*1*cos@ = 740*1*(3/5).
Since T₂ and T₃ are clockwise torques, then they are negative.
So,
T_total = T₁ - T₂ - T₃ = N_wall*4 - w_ladder*1.5 - 740*1*(3/5).
When computing the torques, use either the x-axis or y-axis depending upon the direction of the forces.
If the force(s) is parallel to the x-axis, take the distance along the y-axis, and vice versa.
The distance should always be perpindicular to the line of action of the force.

Thank you very much Fermat. so the key is to take perpendicular distance when calculating for the torque.

 

[Fermat]

That's right. There should always be a right angle between the force and the distance measured.