Recent content by crichardson17

Okay. That makes sense. So, following that, the downward forces (T) on the pulley will be 2(W_A + W_B)?

So, for the second case, where the upward force is equal to the downward force, the FBD would look like this: F_pull = W_A + W_B ∧ ⋅ ∨ T = W_A + W_B so then F_pull would be equal to the tension in the string, wouldn't it? FBD for block A: T ∧ ⋅ ∨ W_A = 196 N (using g=9.81 m/s^2) Since A is...

T is the tension in the string, so... If F is less than the downward forces, the system won't move, and the tension in the string will be equal to the force pulling upwards. If F is equal to the downward forces, the forces will cancel each other out and the system will stay at rest. So again...

Hi all! I'm a student, currently studying geology, math, and physics. Glad to be part of the community!

Homework Statement The blocks of mass 20.0 kg and 10.0 kg are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F is applied to the pulley. Find the accelerations a_A of block A and a_B for block B when F is...