![[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