Physics I: A painter on a chair attached to a pulley.

A house painter uses the chair and pulley arrangement of the figure to lift himself up the side of a house. The painter's mass is 70 kg and the chair's mass is 10 kg. With what force must he pull down on the rope in order to accelerate upward at 0.20 m/s²?

Your first instinct might be to guess that he simply needs to pull down as hard as his weight + the chair's weight plus enough force to accelerate. This is wrong! 

Start by drawing a free body diagram. The painter has a normal force N with the chair (upward), and a reaction force of the chair pushing up on him due to the tension pulling the chair up. He has a downward force equal to his weight. The chair has an upward tension force, a downward normal force from its contact with the painter (N), and its weight.

Fnet(painter) = N + T - W_p = m_pa

Fnet(chair) = T - N - W_c = m_ca

Since acceleration is known, and the painter's pull is equal to the tension force, we set acceleration equal to 0.20 m/s^2 and solve for T. This is a little bit more difficult since the normal force is not known (the normal force in this case is NOT equal to the painter's weight, because the painter is also in contact with the rope pulling downward and the chair is moving upwards!

Solve for both N and T algebraically first.

T = m_pa + W_p - N
N = T - W_c - m_ca

Now plug N into the T equation to eliminate the N variable.

T = m_pa + W_p - (T - W_c - m_ca)
T = m_pa + W_p -T + W_c + mca
2T = m_pa + m_pg  + m_cg + m_ca
T = (m_pa + m_ca + m_pg + m_cg) / 2

Solution: T = 400 N