00:01
In this problem on the topic of static equilibrium, we are told that a person is halfway up a uniform ladder of length 3 .433 meters.
00:08
The person has a mass of 96 .97 kilograms and the ladder has a mass of 24 .91 kilograms.
00:16
The ladder is leaning against a wall and the angle between the ladder and the wall is 27 .3 degrees.
00:23
We'll assume that the friction force between the ladder and the wall is zero, and we want to know the minimum value of the coefficient of static friction between the ladder and the floor that keeps the ladder from slipping.
00:37
Now, if we take the bottom left end of the ladder as our pivot point, we know that the net talk about this point must be zero since the ladder is in static equilibrium.
00:51
So if we take, if we first look at the clockwise talks about the point at which the ladder touches the ground, this is equal to the weight of the ladder times the distance l over 2, sine theta, plus the weight of the person, wp, times the distance l over 2, sine theta.
01:18
Now, the counterclockwise talks about that point, we'll call it tor ccw, is equal to the reaction force r times l times cosine theta.
01:31
Now the net talk is to be zero since we have equilibrium.
01:38
So that means that rl cosine theta is equal to wl times l over 2 sine theta plus the weight of the person, wp, l over 2, sine theta...