Thanks @PeroK.
" There could be any force between the blocks. They could, say, be charged and be electromagnetically repelling each other. They could also be attracting each other."
Suppose there is no such a thing as springs, net electric charge, etc. Only the contact force is present.
Can you...
@BvU
"It is not very handy to take the two blocks as a whole"
If you look carefully I analyzed both, the blocks as a whole and the block with mass ##m_1##. You can do the same for mass ##m_2##, but it is redundant because you can deduce that from Newton's third law.
"Look at the individual...
Thank you for your answer. But, please correct me if I am wrong, are you saying that are friction between the blocks?
If that is what you are saying, this is not the case. The friction is between the platform and the blocks.
Thank you for your answer, but I don´t think is correct. I get an answer for ##a=0##, and for ##a=\mu_{est}g##, and the result can be obtained from Newton's second and third laws of motion.
Taking the masses as a whole, this is the Newton's second law for the system:
$$(m_1+m_2)a=Fr_1 + Fr_2$$
(With ##Fr_1## and ##Fr_2## the respectively friction forces)
Which implies:
$$a=\frac{Fr_1+Fr_2}{m_1+m_2}$$
Newton's second law for the For the ##m_1##
$$m_1a=Fr_1 -N_{21}$$
Then...
Thank you for your questions, BvU.
Yes, the force of interaction between the blocks, or the normal force between them (i.e. the contact force that is perpendicular to the surface of the blocks).
That is the question I can't answer yet. I guess is still 0, but I don't know how to prove it...
Hello, everyone.
This problem is easy if you assume that the friction between the blocks and the platform is the maximum possible. Then the normal force is 0. But how can you show that the normal is 0 even when the friction is less than maximum?
Thank you for your help.
Hello @Mark44, thank you for your reply. I translate my program from spanish and I missed a ""puntos"" where should be changed for "points" , I wanted to edited but I don't know how.
By the way, In mathematica 2energy=2*energy is a valid operation, no need for whitespaces, though integrate2...
I'm trying to solve this problem with Mathematica.
Im not a Mathematica expert, but my program works perfect when vhartree={0,0,..,0}.
This is the program I wrote:
Clear[u, poisson, vhartree]
h = 10^(-2);(*step integration*)
rmax = 20;
rmin = 10^(-30);
Z = 2; (*atomic number*)
points =...
Thank you again, CWatters. Well I know how to solve the problem, and this is out of discussion:
For the pair:
52x3/2!= 78 ways
For the three different cards:
48x44x40/3!=14 080 ways
then: 78x14 080= 1098240 ways to get exactly one pair.
but my problem is: Whats wrong in the procedure i post first?
Im New on this forum, so i hope this is the right place to ask this question. I've tried to solve the problem of finding how many hands are in a poker game with exactly one pair, this way:
I can choose the first card from 52 ways. for the second card I can choose it from 3 ways (to match the...