![[Elevator_counterweight.png]] The two guide rails for the elevator shown above each exert a constant friction force of 100 newtons on the elevator car when the elevator car is moving upward with an acceleration of 2 meters per second squared. The pulley has negligible friction and mass. Assume g = 10 $\frac{m}{s^2}$. a) Draw and label all forces acting on the elevator car. Identify the source of each force. ![[Elevator_Counterweight_1976B_FBD.png]] b) Calculate the tension in the cable lifting the 400-kilogram elevator car during an upward acceleration of 2 $\frac{m}{s^2}$. (Assume g is 10 $\frac{m}{s^2}$.) $\Sigma \vec{F} = ma$ $\Sigma \vec{F} = T-W-2F_f$ $ma = T-W-2F_f$ (400 kg)( 2 $\frac{m}{s^2}$ ) = T - W - 2 (100 N) 800 N = T - 4000 N - 200 N ==5000 N = T== Newton's Second Law The sum of the forces Combine the two equations Substitute in values Solve c) Calculate the mass M the counterweight must have to raise the elevator car with an acceleration of 2 $\frac{m}{s^2}$. $\Sigma \vec{F} = ma$ $\Sigma \vec{F} = Mg - T$ $Ma = Mg - T$ $T = Mg - Ma$ $M(g-a) = T$ M = $\frac{T}{(g-a)}$ = $\frac{5000 N}{10-2 {\frac{m}{s^2}}}$ ==M = 625 kg== Newton's Second Law The sum of the forces Combine the two equations Rearrange to isolate tension Substitute in values Solve