Recent content by engineer5lyfe

For the first case (which I assume you are referring to, as it is the only one with F_in), F_in is simply a force that is required to maintain static equilibrium since no friction is present in the system. It "exists" because it is prescribed by the problem, such that there is static...

I have decided to add an addendum to this post in case others come across it - hopefully it will help build intuition and understanding as to why the normal force changes for different situations. The first example is a "wedge", however the input force required to maintain static equilibrium...

After posting this reply, I now see that F_N is reduced when the wall constraint is removed, because no horizontal force exists, therefore the normal force only has to react to mg, and therefore F_N=mg*cos(theta).

I think I understand - thank you for the clarification. I've included an updated diagram below to check my understanding. So in both cases, F_N=mg/cos(theta)? Why then is F_N=mg*cos(theta) if the wall constraint is removed?

F_x for the second case would be a component of the normal force between the wall and the block, correct? The way that I see it, the sum of the components, F_x and F_y for the second example should equal the normal force on between the block and the wall. Perhaps I am misunderstanding, could...

I have seen a few posts on this subject before, but none have really answered my question. For clarity, I will refer to the 1st example as a wedge, and the second as a ramp (although both are of course inclined planes). With both examples that I outline below, we will assume no friction, and a...