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Solving simultaneous equations F = ma

by Jbreezy
Tags: equations, simultaneous, solving
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Jbreezy
#1
Mar17-13, 02:48 PM
P: 582
1. The problem statement, all variables and given/known data

Luke stands on a scale in an elevator which has a constant acceleration upward. The scale reads 0.97 kN. When Luke picks up a box of mass 20 kg, the scale reads 1.18 kN. (The acceleration remains the same.) (a) Find the acceleration of the elevator. (b) Find Luke's weight.


2. Relevant equations
So I came up with 2 equations one for Luke w/o the added mass one for Luke with.
W_1 = m(g+a) I solve it for m.
m = W_1 / (a+g)
equation 2
W-2-(m+20)g=(m+20)a


3. The attempt at a solution
My strategy is to just solve both equation for m. Set them equal and then just plug and chug and get my my a. The issue is I can't solve the equation!!!!!!!
If my strategy is correct. The units don't work out at all. Apples and Oranges.
So,

W_2 - 20g - 20a / a+g =W_1/ a+g
Multiply both sides by a+g gave
W_2 - 20g -20a = W_1
Cleaning it up for a I got...

W_2 -20g - W_1 / 20 = a
See what I meant about apples and oranges something is wrong. I have been at it for a while I have tunnel vision.
Thanks to whoever helps
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rude man
#2
Mar17-13, 03:12 PM
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Quote Quote by Jbreezy View Post
1. The problem statement, all variables and given/known data

Luke stands on a scale in an elevator which has a constant acceleration upward. The scale reads 0.97 kN. When Luke picks up a box of mass 20 kg, the scale reads 1.18 kN. (The acceleration remains the same.) (a) Find the acceleration of the elevator. (b) Find Luke's weight.


2. Relevant equations
So I came up with 2 equations one for Luke w/o the added mass one for Luke with.
W_1 = m(g+a) I solve it for m.
m = W_1 / (m+g)
What kind of solution for m is that??
Jbreezy
#3
Mar17-13, 03:27 PM
P: 582
I changed it I just meant m = W_1/ (a+g)
That is it. Sorry typo. Good catch.

SteamKing
#4
Mar17-13, 03:37 PM
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Solving simultaneous equations F = ma

I suggest you look at the problem without trying to solve for both weights simultaneously.

You know the scale readings before and after Luke picks up the 20 kg box.
Assuming the accel. in the elevator is the same for both scale readings, and assuming a typical g, then you should be able to find the acceleration of the elevator.
Jbreezy
#5
Mar17-13, 03:41 PM
P: 582
Hello StemKind,
I was actually not just solving for the weights. I know what the weights were. But you are saying I should be able to get this without combining my equations?
Thanks,
rude man
#6
Mar17-13, 04:17 PM
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Quote Quote by Jbreezy View Post
I changed it I just meant m = W_1/ (a+g)
That is it. Sorry typo. Good catch.
Substitute into your W_2 equation to find a. Then get m.
Jbreezy
#7
Mar17-13, 05:04 PM
P: 582
yeah that is what I tried above the units are not correct. That is the strategy but something is wrong with my solution above.
haruspex
#8
Mar17-13, 05:44 PM
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Quote Quote by Jbreezy View Post
W_2 -20g - W_1 / 20 = a
Two things will help here. Include parentheses and either include units throughout or, better, keep everything symbolic until the end. E.g. if mb is the mass of the box (20kg):
(W2 -mbg - W1) / mb = a
... and everything's apples.
Jbreezy
#9
Mar17-13, 05:56 PM
P: 582
Ahh Thank you Haruspex.
Apples are quite good. Especially with peanut butter. I see the issue now.


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