Find Apparent Weight of 85 kg Man on Elevator

[Mitch ODriscoll]

Homework Statement

A man with a mass of 85.0 kg rides downward on an elevator.Initially the elevator accelerates downwards at 1.845 m/s^2 and eventually slows at 1.60 m/s^2 as it reaches the ground floor. Find the mans apparent weight as he starts down and as he reaches the ground floor.

Homework Equations

fn=m(a-g)

The Attempt at a Solution

fn= (85.0 kg) (1.85m/s^2--9.81m/s^2)
=991.1 N

fn=m(a-g)
= (85.0 kg) (1.60 m/s^2--9.81 m/s^2)
=969.85 N

The question I have is in the second part of the question should th value for a be positive or negative?

 

[Doc Al]

This is not quite right: you added instead of subtracted.

It might be clearer if you wrote it as:
$$\Sigma F = F_n - mg = ma$$

So:
$$F_n = m(a + g)$$

But note that the acceleration is downward and thus negative.

 

[Mitch ODriscoll]

So fn=85.0kg ( -1.85m/s^2 +(-9.81m/s^2)
=-991.1N

fn=85.0kg (-1.60 m/s^2 + (-9.81 m/s^2)
= -969.85N
Is this correct?

 

[Doc Al]

No, you're making the same mistake. In the expression "a + g", it's "a" (the acceleration of the elevator) that is negative, not g! (g is just a positive constant = 9.81 m/s^2.)

Always do a sanity check: If the elevator accelerates upward, you'd be squashed against the floor giving a greater apparent weight; if it accelerates downward, you'd be pulled away from the floor, reducing your apparent weight.

 

[Mitch ODriscoll]

fn =85kg (-1.85 m/s^2 +9.81m/s^2)
= 676 N

fn = 85kg ( 1.6m/s^2 + 9.81m/s^2)
= 969.85 N
Is this correct?

 

[Doc Al]

Excellent! The wording of the problem was a bit tricky. At first the elevator accelerates downward at a = -1.85 m/s^2. Then it slows down, which means it accelerates upward at +1.6 m/s^2.

 

[Mitch ODriscoll]

Thanks for the help !