Solving Cartesian Vectors in Mechanics: Expressing Force F(AB) in Cartesian Form

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Robb
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Homework Statement


Hibbeler14.ch2.p92.jpg


Express force F(AB) in cartesian vector form[/B]
Mastering engineering.com says to check my signs but I don't an see an issue. Please help!

2. Homework Equations 3. The Attempt at a Solution

A(0, -.75, 3) B(2cos40, 2sin40, 0) C(2,-1, 0)

r(
AB)= 1.53i +2.04j-3k
r(magnitude)= 3.94

u(ab)= .388i + .518j - 3k
F(AB)= 97i +130j - 190k
 
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Masteringengineering.com says to check my signs. I do not see anything wrong with them. We are asked to express the force F(AB) as a cartesian vector.
 
Your results look fine. I might have rounded the results a bit differently, but your signs are certainly correct.
 
How would you have rounded it. The site is now telling me I have a rounding error. I went through my numbers again and went out to 100 thousandths place. The only change I get is that i=97.1?
 
Now that I've looked again at the problem statement I can see that there's a bit of a question about how many significant figures to use. Many of the distances are given as a single digit with no decimal indicated, but at least one has two decimal places (0.75 m). So do we round to one significant figure or assume that the single digit numbers were really measured to two decimals of accuracy but were shown as single digits for aesthetic reasons on the figure?

My own choice would be to round to two significant figures as you've done in your first post, but there's an argument to be made for rounding to one sig fig!
 
The problem says to round to 3 sig-figs and I entered 97.1i +130j -190k
 
Robb said:
The problem says to round to 3 sig-figs and I entered 97.1i +130j -190k
Okay then. What were your unrounded results to three decimals?
 
97.1475I + 129.53j - 190.485k
 
Robb said:
97.1475I + 129.53j - 190.485k
Did you round or truncate any of your intermediate results to five or fewer digits? I'm seeing slightly different values that could affect your rounding to three figures.

Generally I keep all intermediate values to full precision in calculator memory (or let a software package hold them in variables) and only round for presentation. This prevents rounding and truncation errors from creeping into the significant figure zone.

Before rounding and to three decimals I'm seeing values: 97.316; 129.297; -190.556
 
ok. I got it. Guess I better play it safe and don't round till the end huh. Thanks so much!