Finding Acceleration and Tension, With Friction

[VACA]

Homework Statement

In figure P4.52 (below), the coefficient of kinetic friction between two blocks shown is 0.30. The surface of the table and the pulleys are frictionless. (a) Draw a free-body diagram for each block. (b) Determine the acceleration of each block. (c) Find the tension in the strings.

Homework Equations

Since all accelerations are equal:
-a₂ = a₃ = -a₁₀

Also, a = F/m
And for gravity: F_g=m (in kg)*9.8

The Attempt at a Solution

For the force diagram I know that the 2kg block will have T₁ pulling it to the left and friction pulling it to the right. The 3kg block will have T₂ pulling it to the right and friction and T₁ pulling to the left. And the 10kg ball, sphere, whatever, has T₂ pulling upward and gravity pulling downward.

F_k is 5.88N since F_n is 19.6N (0.3*19.6).
And F_g is 98N.

- (5.88N - T₁) / 2kg = (T₂ - T₁ - 5.88N) / 3kg = - (T₂ - 98N) / 10kg Got rid of my fractions, LCD of 30...

15T₁ - 88.2N = 10T₂ - 10T₁ -58.8N = 294N - 3T₂ Used the first two equations to solve for T₂

2.5T₁ - 2.94N = T₂ Plugged into the last 2 equations and ended up getting...

25T₁ - 29.4N - 10T₁ - 58.8N = 294N - 7.5T₁ + 88.2N

T₁ = 20.906N

The correct answer for T₁ is 17N. I've looked over my work and couldn't find a mistake in my math, though I'm sure there is. I need help figuring out where I messed up.

I hope this isn't just some dumb mistake because I've spent so long on this single problem.

Thanks in advance.

 

[grzz]

25T₁ - 29.4N - 10T₁ - 58.8N = 294N - 7.5T₁ + 88.2N

All is Ok except for a small mistake. Last number is not 88.2 but 8.82.