So it's
For part B, the F=ma equation for mass A is T-mgsinӨ-u(kinetic)mgcosӨ=ma, while for mass B, it is mgsinӨ-u(kinetic)mgcosӨ-T=ma.
So T-829=100a, meaning T is 829+100a. Plugging into the equation for mass B, 391.8-117.7-(829+100a)=50a. So a=-11.1 m/s^2. Is this right?
But then...
Homework Statement
Two blocks connected by a cord passing over a small fictionless pulley rest on an double inclined plane (ie a traingle) with static friction coefficient of 0.5 and a kinetic friction coefficient of 0.4. The mass of block A is 100 kg (sitting at 30 degrees), while the mass of...