Ok, the friction in the pulley is going to be (by capstan equation & assuming block_A will tip over): ##W=T*e^{\frac{2}{\pi}*\frac{\pi}{2}} \rightarrow W=50*e=135.91N##
Please, tell me this is correct :')
Note: I edited the angle which I substituted in the calculations as Mr. @Steve4Physics had...
Some hinge from another problem
Ok, so that's my mistake it should tip first because the force required for making tip is much smaller
Something circlish, more precisely to let the angle which the rope is touching the pulley over fit in the capstan law well, as the coefficient is multiplied by...
Ok, I'll do
Here's what I got:in the first case (Block A tips):##\sum{}^{} M_{right\ bottom\ corner} = 0 \rightarrow 200*1=4*T\ (tension\ in\ the\ rope\ connected\ to\ the\ block) \rightarrow T=50N##
The second case (block A slips alone): ##\sum{}^{} F_x =0 \rightarrow 0.5*200=T \rightarrow T=...
First, I assumed that the tension in the rope connected to the block A equals the static friction ##\sum{}^{} F_x =0 \rightarrow T=N_A*0.5=100N##, then the W weight or force equals to the tension in the pulley and the tension T ##W=100+\frac{2W}{\pi} \rightarrow W=275.2N##
It's the first problem...
Zero, isn't it?
Maybe I'm asking because the model answer for this problem used the ##N_A## we got earlier in the calculations, so that made me feel that there's sth wrong.
The first thing I did is to get the value of ##N_A## by the equaiton of ##\sum{}^{} M_B=0 \rightarrow 650*0.4+450*0.45+400=N_A*0.6 \rightarrow N_A = 104.2N## This is the first.
The magnitude of the reaction at B: ##\sum{}^{} F_y=0 \rightarrow B_y=450+650,\sum{}^{} F_x=0 \rightarrow...