The following problem is from my physics textbook: A crate slides down an inclined right-angled trough. The coefficient of kinetic friction between the crate and the trough is [itex]\mu_k[/itex]. What is the acceleration of the crate in terms of [itex]\mu_k[/itex], [itex]\theta[/itex], and g? The problem was accompanied by a drawing which I have tried to reproduce and included as an attachment. The answer I come up with when I try this is [itex]g(sin \theta - \mu_k cos \theta)[/itex], which differs from the book's answer in that the book's answer has a factor of √2 sitting in front of the cosine. I think I understand how to do problems involving friction and inclined planes in general, but I do not understand why it matters that the crate is in a right-angle trough, or in a trough at all for that matter. I recognize that a factor of √2 would result from an isosceles right triangle, or from a 45-degree angle, which you could certainly form out of the many right angles present in the problem, but I don't see how any of them would affect the final answer. I have tried to calculate the force the crate exerts on each side of the trough, and then take the vector sum of those two forces to find the net normal force acting upon the crate. When I did this, I found the result to be equal in magnitude to the gravitational force acting orthogonal to the trough - in other words, right back where I started. Perhaps I made a mistake here? How can I correctly solve this problem?