# Help to understand

1. Nov 26, 2005

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.
(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

2. Nov 26, 2005

### 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.

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 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: Nov 26, 2005
3. Nov 26, 2005