Question

An elevator and its counterweight are connected by a mass-less cable over a frictionless pulley of negligible mass. The counterweight has a mass of 1,000kg and the elevator with four passengers has a mass of 1,150kg. Calculate the acceleration of the elevator and the tension in the cable.

Answer 1

\[\sum_{?}^{?}F _{E} = T - mg _{e} = m _{e}a\]

\[\sum_{?}^{?} F _{C} = T - m _{c}g = m _{c}a\]

Answer 2

draw it :)

Answer 3

\(m_{E} g - T = m_{E} a, \qquad T - m_{cw} g = m_{cw} a\)

notice the difference

Answer 4

So make each equation equal to each other?

Answer 5

This is a 1-D problem. Up and down motion, sure, but this is 1-D physics.

i don't know if that helps but look at the number you expect to get for \(a\), the acceleration.

your and my equations have really different consequences in terms of the solutions.

Answer 6

I see, since the tension for the elevator is allowing for it to go down... we expect a negative number for our acceleration.

Answer 7

since this is a 1-D problem, acceleration is just a number on the number line.

the fact that it is upwards on the RHS and downwards on the LHS is irrelevant.

Answer 8

so the trick is getting the signs right

as i said, look at your equations and the ones i suggested

solve them to see how they differ

Answer 9

Can you explain how you derived your equations?

Answer 10

tx genius

Answer 11

:)

Answer 12

Rederiving them...

\[T = m _{E}g - m _{E} a\]
\[T = m _{cw}g + m _{cw}a\]

Answer 13

This would work?

Answer 14

yes. totally

Big QU is: can you intuit it?

Answer 15

\[a = \frac{ 9.8(1150 - 1000) }{ (1150 + 1000) }\]

\[a = 0.68\]

Answer 16

Made the tension equations equal to each other then solved for a

Answer 17

\(\checkmark\)

Answer 18

\[T = 1150(9.8) - 1150(0.68)\]

\[T = 10,488 N\]

Answer 19

\(\checkmark\)

Answer 20

Nice, thank you

Answer 21

TU2 :)