I don't know the word in english but by removing I mean when you draw the object and the forces (what I learned in class) you can sometimes remove certain objects and then draw the force it makes on the other object. Hope you understand what I mean.
If the rod CB was perpendicular for instance...
I also have one more question about the way I solved it the first time. I can only remove the rod CB and write a force F like that if the rod CB was perpendicular right?
Wow I didn't know it was this complicated haha. Thanks for clarifying it to me, now I understand how to think about these kind of questions. Thanks for the help!
I actually didn't, just assumed the force would be that way if you removed the bar CB. But now that you write this I realize that method 1 only works if bar CB was perpendicular right?
Does it mean the second method is the right one?
Sorry but my english is not great. I have no idea what you...
I solved this with two methods as you can see down below in the picture. Which is the correct way?
I remember I learned in class you could use both ways but why am I getting different results?
It was a long time ago I did these kind of problems so I’m a bit rusty. The only thing I can think of is divide it up to two parts one x and one y.
In y the acceleration is sin(a)*9.82? Then put that in the equation and solve for t.
In x the there is no acceleration so the formula is x=V0*t, I...
Thanks for clarifying this, now this question seems a lot easier. By the way do you know if there is anyway you can mark this thread as complete or solved? Or do you just leave it be when the homework is solved?
And again thanks to you and TSny for helping me!
Changed that sign to + and now I got the correct answer! Thanks you so much.
I could only solve it with my method by placing in the equation for FB and NB into my last equation. But you wrote something about substituting into the first equation, how? Would it be easier to solve this kind of...
Look on picture two for my equations, I’m pretty sure my first three equations are correct as it’s the same on the answer paper I got. Answer paper only shows the three equations and correct answer so I have no idea how they got to it.
When I put the equations togheter And solve out ”h” i get...
Ohh now I get it, if the force Q had been straight above point B then I would ’t need to take the horizontal force into account? Did I understand it right?
Alright I understand what you mean now and got the answer to 877N, confirmed with a friend and he got the same.
Now I understand you need to take into account the vertical and horizontal forces on a force that is skewed (can’t find another word to describe force Q).
Can you please explain to...