TP 11.20

Given: n_L = 290 N

            n_R = 122 N

            L = 2.50 m

(a)  Find m - ?

From the condition of translation equilibrium:

(n_L)_x = n_L cos (90^o) = 0

(n_L)_y = n_L sin (90^o) = n_L  = 290 N

(n_R)_x = n_R cos (90^o) = 0

nF_R)_y = n_R sin (90^o) = n_R = 122 N

  w_x = w cos (-90^o) = 0

w_y = w sin (-90^o) = - w = - m g

  SF_x = 0 

SF_x = 0    (290 N) + (122 N) – mg = 0

mg = 412 N

m = 42.0 kg

(b) Find x -?

From the condition of rotational equilibrium:

  t_nL  = 0

t_nR  =   (122 N) (2.50 m) = 305 N-m 

t_w =  - (412 N) x 

St = 0 

(305 N-m) –(412N) x = 0

  x = 0.740 m