A rock climber of mass 55 kg is hanging suspended from a rope tied to another climber of mass 65kg on a horizontal cliff ledge.if the coefficient of kinetic friction (mu) between climber on the ledge and the kedge is 0.45; what is the net acceleration of the climber on the ledge?

Question

aaronq · Answer

|dw:1405478790563:dw|

Newton's laws for climber on cliff

x-direction

$\sf \sum F_x=F_{pull}+f=ma$

$\sf \sum F_x=m_1g-f=(m_1+m_2)a$


y-direction

$\sf \sum F_y=n+F_g=0$

$\sf \sum F_y=n-m_2g=0\rightarrow n=m_2g$


friction = $\sf f=\mu_kn=\mu_km_2g$

sub into x:  $\sf m_1g-\mu_km_2g=(m_1+m_2)a$


$\sf a=\dfrac{g(m_1-\mu_km_2)}{(m_1+m_2)}=2.1~m/s^2$