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Catchingupquickly

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## Homework Statement

An elevator containing three passengers (with a mass of 72 kg, 84, and 35 kg, respectively) has a combined mass of 1030 kg. The cable attached to the elevator exerts an upward force of 1.20 x ##10^4## N but friction opposing the motion of the elevator is 1.40 x ##10^3## N.

a) draw a free body diagram of all the forces acting on the elevator.

b) Calculate the net acceleration of the elevator and its passengers.

c) Draw a free-body diagram of all the forces acting on the 35 kg passenger.

d) Calculate the force normal acting on this passenger.

e) What velocity will the elevator have 12.0 seconds after the passengers have entered the elevator?

## Homework Equations

## acceleration = \frac F m##

FN = mg - ma

v = v0 + (a)(t)

## The Attempt at a Solution

[/B]

So 1.20 x ##10^4 N ## = 12 000 N [ up]

and 1.40 x ##10^3 ## = 1400 N [down]

Fnet is thus 12 000 - 1400 = 10 600 N

g = 9.80 ##m/s^2##

a) I have lackluster drawing skills on this forum so a word picture: An arrow up from a central point. Top of the arrow labelled Fnet. Two arrows down from the central point on the same line, 1st arrow labelled Fg and the bottom, last labelled Ff

b) acceleration = Fnet / m

= 10600 N / 1030 kg = 10.29126213592233 ##m/s^2##

c) Two arrows going up on the same line from the central point. The first arrow labelled Fnet and the top labelled FN. The arrow moving down from the central point is labelled Fg

d) ##FN = mg - ma

\\= 35 kg (9.80 m/s^2) - 35kg (10.29126213592233 m/s^2)

\\ = 343 N - (- 360.194174757278 N)

\\ = 703.194174757278 N##

or 703.2 N or just 703 N

e) ##v = v0 + (a)(t)

\\ = 0 + (10.29126213592233 m/s^2) (12.0 seconds)

= 123.495145631068 m/s##

or 123.5 m/s or 124 m/s

I'm not feeling confident about these answers. If I'm right this is a kick @SS, potentially deadly express elevator. Which kinda makes me think I missed something somewhere.

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