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## Homework Statement

This is a two part problem I figured out the 2nd part but not the first.

a) A 7.3 m ladder whose weight is 364 N is placed against a smooth vertical wall. A person whose weight is 810 N stands on the ladder a distance 2.7 m up the ladder. The foot of the ladder rests on the floor 3.577 m from the wall.Calculate the force exerted by the wall. Answer in units of N.

b) Calculate the normal force exerted by the floor on the ladder. Answer in units of N. 1174 N

## Homework Equations

Forces:

x-axis: F

_{k}- N

_{Wall}= 0

y-axis: N

_{floor}- W

_{person}- W

_{ladder}= 0

Torque:

N

_{wall}x h - W

_{ladder}x [tex]\frac{h}{2}[/tex]cos[tex]\theta[/tex] - W

_{person}x (2.7 x cos[tex]\theta[/tex]) =0

## The Attempt at a Solution

[tex]\theta[/tex] = cos

^{-1}([tex]\frac{3.577}{7.3}[/tex]) = 61

tan[tex]\theta[/tex]61=[tex]\frac{h}{3.577}[/tex]=6.45

3.65cos61=1.77

2.7 x cos61 = 1.31

N

_{wall}x 6.45 - 364 N x 1.77 - 810 N x 1.31 =0

N

_{wall}x 6.45 -644 - 1061 = 0

N

_{wall}x 6.45 = 1705

N

_{wall}= [tex]\frac{1705}{6.45}[/tex]

N

_{wall}= 264 N

Its wrong. What am I doing wrong?