Recent content by YehiaMedhat

The question is: A hemispherical shell of radius ##4.0 cm## is located at the region ##z > 0.0## centered at the origin. It carries a surface charge distribution of density ##\rho = 2.0\times 10^{-6} C/m^2## in the region ##y > 0.0## and a surface charge distribution of density ##\rho =...

Is that also wrong? How the W on the other sould be greater so it makes sense to put on that side.

It was stated to use g as approximation, it's not me. But the rest now is correct?!

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...

This what the numbers got me, so, what's wrong?

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=...

I tried it but it will only affect the numbers, won't it?

This seems much more complicated than what I thought But this should be solved simply my curriculum has no utterance about that stuff

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...

Ok, that makes sense now, so it'll be##\sum{}^{} M_B=0 \rightarrow 0.45P=400-650*0.4 \rightarrow P=311N##

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 picture for the problem its self And I did refrence the forces by the arrows and the name of the points like##N_A##

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...

That's great. 👍