Question

An elevator weighing 20 000 N is supported by a steel cable. What is the tension in the cable when
the elevator is being accelerated upward at a rate of 3.00 m/s2? (g = 9.80 m/s2)

Answer 1

There are two forces acting on the elevator. There is, of course, the force of gravity pulling the elevator downward. There is also the force of tensionexerted by the cable on the elevator. The tension force from the cable is pulling the elevator up.

We know all about the force of gravity since it is equal to the weight of the elevator (it is given to us) and acts downward.
But we only know the direction, not the magnitude, of the force of tension.

We do, however, know something else…..Newton’s 2nd law.
Net Force = mass * acceleration,
F = ma

We know what the elevator’s mass is since we know the elevator’s weight (weight = mass * gravity, so mass = weight / gravity). We are given what theelevators net acceleration is (3.00 m/s^2 upward). So we can determine the net force.

Assuming gravity, g, = 9.80 m/s^2, then

The mass of the elevator must be,
m = (20,000 N) / (9.80 m/s^2) = 2041 kg
so now,
F = (2041 kg) * (3.00 m/s^2) = 6122 Newtons directed upwards

This is the net force.
The net force can also be found as the difference between the upward and downward acting forces,
F_net = F_up – F_down

The only downward acting force is that of gravity and the only upward acting force is the force of tension. So,
F_net = F_tension – F_gravity

We can solve for F_tension now as,
F_tension = F_net + F_gravity
F_tension = (6122 N) + (20,000 N)
F_tension = 26,122 Newtons, directed upward.

So the force of tension in the cable is about 2.6 E4 Newtons, causing the elevator to accelerate upwards at a rate of 3 m/s^2.

Answer 2

When an object moves upwardswith acceleration 'a' , effective value of g isg'=g+ aGiventhata=3.0m/s²g'=9.8+3.0=12.8m/s²Mass ofelevator=weight /g=20000/9.8=m=2.04* 10³kgTension incable=effectiveweight=m *g'T=2.04*10³*12.8=26122.45N