Question

How far along the ladder can a 750 N person climb up the ladder?

Answer 1

We will ahve the following:

Forces on x:

$\sum ^{}_{}F_x=0=\sigma-n_1\colon\sigma=n_1$

Forces on y:

$\sum ^{}_{}F_y=0=n_2-200N-750N\colon n_2=950N$

Since it is a stable system we will have:

$n_1=\mu n_1$

Now, we will determine the torque at the bottom of the ladder:

$\sum ^{}_{}\tau_{\text{bottom}}=0-(200N)(3m)\cos (56)$

Also:

$-(750N)(d)\cos (56)$

&

$+(n_1)(5m)\sin (56)$

Finally:

$0=-600\cos (56)-d\cdot750\cos (56)+(285)(5)\sin (56)\Rightarrow-d\cdot750\sin (56)=600\cos (56)-(285)(5)\sin (56)$$\Rightarrow d=\frac{(285)(5)\sin (56)-600\cos (56)}{750\cos (56)}\Rightarrow d=2.01686584$$\Rightarrow d\approx2.02$

So, the distance along the ladded that a 750N person could climb in the ladder would be approximately 2.02 meters.