Recent content by flanders

SammyS: I have the same results as you have =). I have written N=Gy+Fy, so that they will have their "correct" direction... I think! tiny-tim: I have done it the same way as SammyS: I did then put N into the x-equation: 0.8F-0.6mg-0.25(0.8mg+0.6F)=0 calculate and it would be...

Hey Tiny-Tim! I've tried to set up two equations (X and Y-direction) and to solve it without results :(. Could you please explain the quadrilateral-method in short terms? I would really appreciate it =). It should also be said, that the answer is to be given in m*g. For instance 0.8mg. The...

I'm sorry - I didn't specify the direction of the force, but we're supposed to find the horizontal force. Thank you very much, it was helpfull :)

Adding all the "main" vectors gives us a triangle with angles of 90, 36.9 and 53.1 degrees. As seen on the drawed picture attached, I have: N\; =\; \frac{G}{\mbox{C}os\left( 36.9 \right)}\; =\; 1.25G F\; =\; N\cdot \sin \; \left( 36.9 \right)\; =\; 1.25G\cdot \sin \; \left( 36.9 \right)\...

Great! Thank you - I think I understood a little bit more! :) Now I used tan to find F, and then Sin to find N. As a result I got F = 4,30m. Is it possible to do something with the m..? I don't think we are able to find it with that sort of information we've got?

I got curious in how you do the vector triangle? You just put the vectors one after another and then you'll find out that you get back to the starting point - therefore the velocity must be constant? And what will the different angles look like? Hm... I'm sorry, but I don't understand that much...

Homework Statement Hi. My task is to find the Force F, so that velocity is constant. I know that the velocity is constant, therefore the sum of all forces must be zero! On the image I have used vector decomposition (the black lines). The orange ones are the vectors. I know that the...

Oh my! :redface: THANK YOU SO MUCH! You don't know how confused I was over this 'easy' task... God bless you!

Okay, here is my solution: Particle 1: a = -4 v0 = 5 x0 = 0 Particle 2: a = 4 v0 = 0 x0 = 1,28 Therefore; Particle 1 to the left, particle 2 is on the right: 0 + 5t + 0,5*(-4)t^2 = 1,28 + 0t + 0,5*4t^2 5t - 2t^2 = 2t + 1,28 -4t^2 + 5t - 1,28 = 0 This leaves...

Homework Statement Hi. I've had some trouble solving this exercise, and I simply can't find the right solution. Hopefully you might help me out...? Here's the task: A plane (not airplane) has an inclination that allows a particle sliding on the slippery slope, and has an acceleration of 4.00...