Thank you very much. Sorry for late reply but I had been busy a lot last days. The problem I got is indeed very easy, but I needed to solve it on various ways, by that I mean to prove all of my results in all sort of ways, mostly vectors.
Im sorry, but I noticed what you wrote right now. So when I calculate all of this in the way we discussed here, what else should I do to give the force in component form and determine the magnitude of that force. The magnitude as you said is the same as the magnitude of acceleration, but how...
Yes, I forgot to say that what I meant by force vector is radius vector from XY(0,0) to XY(2,5). You definitely made me aware to be careful with the terms and be more specific so, thanks again.
So, just to make myself sure before I go and explain my homework in class and get graded, direction of ##\vec F## will be pointed from XY(0,0) to ##\vec a = 2\vec i + 5\vec j m/s^2## ? I mean, that is the only way that seems logic because, the ##\vec a## depends on ##\vec F##, but not the opposite.
In future I won't remove template headers. Can you please define the ''magnitude'' how is it different from resultant force. When I translate magnitude, I get the same translation as size. How is it different from what I get by calculating the force with that equation? I am sorry if my questions...
By that you mean that text that was written by 1.2.3. numbers? If it is that, than it's at my knowledge. Hmm, my task asks for resultant force ##\vec F## and it's ''amount'' . I am translating it literally from my language as it says in the task. So with L2-norm I can find the amount of that...
1. Body of mass m = 3kg is accelerating with acceleration of ##\vec a = 2\vec i + 5\vec j m/s^2 ##. I need to find the resultant force ##\vec F## and the force's value.
2. Of course, the first equation that came to my mind is ##\vec F = m * \vec a##
3. I came with a possible solution for this...
Hello, I am a first year Department of Physics student. I am relatively new to Physics and I am just discovering the beauties of this science. I found about this community on another forum thread so, I am here to learn some things, ask for help when in need, and of course, help others if can.