How Does a Ladder in Equilibrium Affect the Forces on a Window and Ground?

[B-80]

A 80 kg window cleaner uses a 16 kg ladder that is 5.5 m long. He places one end on the ground 2.1 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 2.9 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder did not slip.

(a) Find the magnitude of the force on the window from the ladder just before the window breaks.Okay, so I know that since it is in equilibrium the sum of the torques must be 0. Therefore I can say $$\tau = 0$$ also $$\tau = \r F sin \theta$$

so $$\tau = \r F sin \theta ladder + \r F sin \theta man - \r F sin \theta window = 0$$

my book told me to set up the equation at the bottom of the ladder so I don't need to include the pavement, anyway when solving this, I get $$\tau = (1.1)(80)(9.8)sin(67.6) + (1.55)(16)(9.8)sin(67.6) - window = 0$$

with that I get 1021.29 which is wrong, any ideas?

Also there is an r in all those equations, but I don't get the formatting, so I can't put it in, but I got the radius of the guy and the ladder in there. also I once solved for 2.9 as the distance he was up the ladder in the y direction, also didn't work

(b) Find the magnitude and direction of the force on the ladder from the ground just before the window breaks.Dind't start this yet, as I am not done part A

 

[B-80]

Bump, any help is really appriciated

 

[ogb p]

First, let's clarify the given information. The window cleaner has a mass of 80 kg and the ladder has a mass of 16 kg. The ladder is 5.5 m long and is placed 2.1 m from the wall. The window cleaner is 2.9 m up the ladder when the window breaks. We are neglecting friction between the ladder and the window, and assuming that the base of the ladder did not slip.

To solve part A, we need to find the force exerted on the window by the ladder just before the window breaks. Since the system is in equilibrium, the sum of the torques must be zero. We can set up the equation as follows:

\tau_{ladder} + \tau_{window} + \tau_{man} = 0

We can also use the equation \tau = \r F sin \theta to calculate the torque at each point. Since we are setting up the equation at the bottom of the ladder, we can ignore the torque from the pavement. Therefore, we have:

\tau_{ladder} = (1.1)(80)(9.8)sin(67.6) = 1021.29 Nm

\tau_{window} = -\r F_{window}sin\theta_{window} = -\r F_{window}sin(90) = 0

\tau_{man} = (1.55)(16)(9.8)sin(67.6) = 248.11 Nm

Substituting these values into our equation, we get:

1021.29 + 0 + 248.11 = 0

Therefore, the force exerted on the window by the ladder just before the window breaks is 1021.29 N.

To solve part B, we need to find the force exerted on the ladder by the ground just before the window breaks. We can use the equation \sum F_{y} = 0 to find this force. Since the system is in equilibrium, the sum of the forces in the y-direction must be zero. Therefore, we have:

F_{ladder} - F_{ground} = 0

We know that the weight of the ladder is given by W = mg, so we can substitute this in for F_{ladder}. We also know that the weight of