Recent content by Larz31

Yes it is homework, due in about 15 minutes :P The only thing I can think of and have tried is the above, resulting in 18 choose 3. I'm not sure if my thinking is correct in saying that if the sum of the initial groups is -1, then I can add one to the total... I'm mainly trying to apply what I...

Ah thanks, Also I left this one out initially: x + y + z + t = 14, where x, y, z, and t are integers that is: i) x >= 3, y>= -1, z>= -4, t>=1 Would this just be 18 choose 3? 3 + -1 + -4 + 1 = -1, so add 1 to 14 = 15 15 + 3 (The dividers) = 18 and then choose 3...

I probably should have included this link: http://mathforum.org/library/drmath/view/68607.html Your answer does seem to be a bit small though, only 68 solutions with all odd integers but it seems right, thanks. EDIT: Does your answer take into account negative numbers as well? And I believe...

Homework Statement Find the number of solutions to the equation x + y + z + t = 16, where x, y, z, and t are: i) odd integers ii) even integers Homework Equations N/A The Attempt at a Solution I know the total number of solutions is 19 choose 3. So I'm thinking I can obtain the...

There are two things that are 20m apart. There is a 40m wire. There is a force of 375N acting down 15m from one end of the wire. That is what I have gathered. I do not see how I can use any of this data in conjunction with each other.

Hmm, I believe the problem I'm having is seeing the picture. I don't really understand what being suspended between two points 20m apart means exactly. Does this mean that the wire is held in place by two things 20m apart ANYWHERE on the wire?

Homework Statement A steel wire 40m long is suspended between two fixed points 20m apart. A force of 375N pulls the wire down at a point 15m from one end of the wire. State the tension in each part of the wire. Known Data Steel wire - 40m Two points that are holding the wire up is 20m apart...

Ah I see now, so then for part a) a(m1 + m2) = Fgx - Ff1 - Ff2 a = [ (2)(9.8)(sin60) - (0.14)(9.8)(0.8 - 2.0) ] / (0.8 + 2.0) = 6.65 m/s/s It is sin60 right? I'm not too sure if that's it, for some reason I just know to always use sin. For part b) then a(m1 - m2) = 2T + Ff2 - Ff1 - Fgx...

I'm confused, if m1a = T - Ff then 0.8a = T - (0.14)(0.8)(9.8) in the x direction. This means that I would need the tension to solve for a and vice versa.

well for the 0.80kg mass, we have: Fnet = Fg - Fn = 0 in the y direction Fnet = Fgx2.0 - Ff (0.8)a = (2)(9.8)(sin60) - (0.14)(0.8)(9.8) Or so I believe... for the 2.0kg mass we have: Fnet = Fgx - Ff + Fgy - Fn (2.0)a = (2)(9.8)(sin60) - (0.14)(2.0)(9.8) I have a feeling I am doing...

Homework Statement This is a problem off of http://www.physics247.com/physics-homework-help/pulley.php It is question number 6 but with an additional part and slight modification by my teacher. I will copy and paste it here but it may be best to view the link for the diagram. Two blocks, one...