Never mind. i figured it out. the diagram didn't even come out right anyways.
Tension(T)=mu_static(μ_s)* mass(m)* gravity(g)
T=.65(20)(9.8)= 127.4 N
T=m(g)
127.4=m(9.8)
m=13 kg for block A
Block B in the figure(Figure 1) rests on a surface for which the static and kinetic coefficients of friction are 0.65 and 0.40, respectively. The ropes are massless.What is the maximum mass of block A for which the system is in equilibrium? (Consider the state of the system shown at the figure only. The system is not in equilibrium when the blocks start to move.)
Express your answer to two significant figures and include the appropriate units.
20kg
Block /45 Degrees
B--------/-;---Rope
_______ ' ;
; ' ;
; ' ;Wall
Wall ; A ;
; Block
i would like to link the picture but i cant so i figured i would draw the figure to the best of my ability. the dashes - /and ' leading from B down to A is the rope.Walls are the :
3 answers
123
123