# Help to understand

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

I dont know how they get the equation especially the underlined coefficients.
My problem is that I dont 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

Homework Helper
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.

Nwall is at A and is normal to the wall. Hence the perpindicular distance of Nwall from the ground, and hence also from the point B, is 4m, giving T1 = Nwall*4.
The torque of wladder about B is the force wladder times the perpindicular distance from B to the line of action of wladder, which is 2.5m*cos@, giving T2 = wladder*2.5*(3/5) = wladder*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,

Ttotal = T1 - T2 - T3 = Nwall*4 - wladder*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.

Last edited:

Fermat said:
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.
Nwall is at A and is normal to the wall. Hence the perpindicular distance of Nwall from the ground, and hence also from the point B, is 4m, giving T1 = Nwall*4.
The torque of wladder about B is the force wladder times the perpindicular distance from B to the line of action of wladder, which is 2.5m*cos@, giving T2 = wladder*2.5*(3/5) = wladder*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,
Ttotal = T1 - T2 - T3 = Nwall*4 - wladder*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

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

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving