Yep. That's the integral I got after being more careful with the projections of ##\mathrm{d}\boldsymbol{E}##. ##\mathrm{d}E\sin\phi## just gives the component of the field parallel to the ##xy##-plane. From this one the final component on the ##x## axis has to be found...
Yep. That's the answer after verifying with the textbook, but I'm not sure what I'm doing wrong with the first result.
It looks like ##\mathrm{d}E_x = \mathrm{d}E_x\cos\theta##, so the integral would be something like this:
##\displaystyle E = \frac{\sigma}{4\pi \varepsilon_0}\int\limits_{\pi...
Homework Statement
Calculate the Electric Field generated by a hemisphere of radius ##R## (and uniform charge density ##\sigma##) at its center.
Homework Equations
##\displaystyle E = \frac{Q}{4\pi \varepsilon_0 r^2}##
The Attempt at a Solution
I think I'm doing some mistakes with the...
I would like to try this case:
In my new diagram the forces are:
- On ##m_1##:
##F_1 = T_1 - w_1##
- On ##m_2##:
##F_{2(y)} = w_1 - (N\cdot \cos(\theta) + T_2\cdot \sin(\theta))##
##F_{2(x)} = N_2\cdot \sin(\theta) - T_2\cdot \cos(\theta)##
- On the pulley:
##I\alpha =...
Sorry for my late reply.
Here is my diagram:
True. From the fact that the pulley needs a net torque for it to rotate and hence that both tensions must not be in balance, the acceleration is not the same.
Well, I must start again. This is what I think:
- Forces on ##m_1##:
##F_1 = T_1 -...
Homework Statement
NOTE: This is a problem I found and the question is actually the part B of it. That of ''massless pulley'' I suppose was dealing with the part A (which I don't know what question was), so please don't consider that (:.
Homework Equations
##F_{net} = ma##
##\tau =...
Aah, I see that the statement of the paragraph above does not contradict what I learned about acceleration, as I was thinking.
Now it's much clearer.
Thank you :)
I'm beginning the chapter of Rotational Motion and Angular Momentum and it says the following which got me confused:
Source: http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@8.32:68/College_Physics
When I was introduced acceleration at the beginning, it was stated that an...
The picture I had in mind is that that I wrote before: both boxes on ##M##, touching each other with a force ##m_1g## to the right.
Maybe this was making me miss the fact that the horizontal force on ##m_2## does not affect the vertical force on ##m_1## (under the absence of friction or any...
There is no basis, honestly. Since the only force acting on ##m_2## is ##T##, the acceleration is ##\frac{T}{m_2}##. I can't argue against that, but I still can't throw out the other idea from my head either.
But what if I don't applied the force ##F##? what would the acceleration of ##m_2##...
Now I think I can print the thread and stick it to the refrigerator :) . I almost lost my mind with this problem.
I would like to understand why the reasoning I went through, after having skipped the first acceleration I got, is actually wrong.
I wanted to know the acceleration of ##m_2## in...
Since my system are the three boxes (and there is no friction on the floor either), my force ##F## has to be:
##F = \small{(M + m_1 + m_2)}\frac{m_1g}{m_2}##
It doesn't look that bad, does it?
On ##m_2##, the vertical forces are in balanced. There is no friction between the surfaces so only the tension, pulling it to the right, is my ##F_{net}## there:
##F_{net} = T##
##m_2a = m_1g##
My acceleration has to be: ##\frac{m_1g}{m_2}##.
The tension I want, right: ##m_1g##?
Since the force of gravity due to ##m_1## is transmitted undiminished to ##m_2## through the string, there is only one: ##m_1g##.
Ok. |:
The ##F_{net}## I want has to be zero.
Tension has to be equal to ##m_1g##.
|:
What I thought was that ##m_1## had indeed a vertical acceleration ##g##, then I concluded that this was not its net acceleration, because of the string and ##m_2## attached to it, but that acceleration I've written before.
It seems that I just want to obtain an answer but I actually want to...