Recent content by blondii

I did the drawings using Adobe Illustrator

Thanks for confirming grzz. Much appreciated

Please find attached also my revised answer on PDF and confirm if I am on the right track now. Thanks

Thanks for your response haruspex. I have reworked the problem with tensions added as specified. See my new attached diagram. Here is my new working out also. Please confirm if my answer is correct this time or any other suggested approaches. Thanks: Let: fk = Kinetic Friction uk =...

Homework Statement A block of mass m1 on a rough horizontal surface is connected to a ball of mass m2 by a light weight cord over a light weight, friction-less pulley, as shown in the figure below. A force of magnitude F at an angle θ with the horizontal is applied to the block as shown. The...

Thanks for the reply who. Much appreciated. Cheers

Just a little correction. After expanding the diff equation I get xr(rx-1+1)(3+r) = 0 The answers are r = -3, r = -x or r = undefined. Please advise if I am on the right track. Thanks

The Question: Find the value of r such that v = xr is a solution of xd2v/dx2 + (x+4)\frac{dv}{dx} + 3v = 0 My Solution: After finding the 1st and 2nd derivative of v and substituting into the equation to equat to zero and look for r, I get the answer r =-3. I also get a root that is...

Ok after considering the equation again I came up with the eigen vector <2,1,1,1>. Would this be acceptable since it seems to be linearly independent and sill satisfies the equation (A- I)v= <0, 1, 1, 1>?

I understand the zero vector can't be used, but after solving the equation earlier that seems to be the only possible solution because all other vectors that will eventuate out of solving that equation still won't be linearly independent to <0,1,1,1>. I tried inputing the matrix into...

With these new corrections there will be repeated roots of 1 which will also generate linearly independent eigenvectors one of which is the zero vector [0,0,0,0].

So after calculating would my eigen values be: λ1 = 1 λ2 = 1 λ3 = (\frac{1}{2})(-1 + \sqrt{3}i) λ4 = (\frac{1}{2})(-1 - \sqrt{3}i) Is this correct now?

Thanks HallsofIvy I realize now where my mistake was. I was not calculating the determinant of the characteristic equation properly and always kept stoping after the first phase (I guess that's what you get for staying awake too long). Thanks guys, your input was much appreciated.

Thanks Simon I will have a look into that.

I have a matrix and can't seem to get my head around finding all the eigen vectors. The matrix is A: (1 0 0 0 1 0 0 1 0 1 0 0 0 0 1 0) I got the eigen values as: λ1 = 1, λ2 = λ3 = λ4 = 0 For λ1: The eigen vector V1 is (0, 1, 1, 1). For λ2 -> λ4: The only eigen vector I...