Recent content by DeviantSOP

Ha, can't believe I missed that. Thanks a lot. Amending my solution. [B]My solution so far[B] B_torque - A_torque = mgb - mga = mg(0.7)-mg(0.1) =0.6mg Torque = Iα I = m(r_a^2 + r_b^2) 0.6mg = m(r_a^2 + r_b^2) * α α = 0.6g / (r_a^2 + r_b^2) a = α * r Acceleration of a =...

Not really sure what's wrong with them. I used net torque = Iα and then I = mr^2 and also α = a * r . As I said earlier, i have no idea what I am doing.

My solution so far B_torque - A_torque = mgb - mga = mg(0.7)-mg(0.1) =0.6mg Torque = Iα I = m(r_a^2 + r_b^2) 0.6mg = m(r_a^2 + r_b^2) * α α = 0.6g / (r_a^2 + r_b^2) α = a * r a = α/r Acceleration of a = (0.6g / (r_a^2 + r_b^2) ) / 0.1 Acceleration of b = (0.6g / (r_a^2 + r_b^2)) /...

Two particles(A and B) each of mass 10kg are attached to a massless rod on each on opposite ends of the rod. The rod is placed horizontally on a fulcrum. Particle A is 10 cm away from the fulcrum. And particle B is 70cm away from the fulcrum. The rod is now released. What is the magnitude of...