What went wrong with finding the resultant?

In summary: I'm not trying to sound arrogant but I want to exercise the mind God blessed me with.But don't get me wrong. If I have to calculate arctan(0.314573) in a hurry, forget Taylor Series. I'll use the calculator.
  • #1
Sting
157
2
I should slap myself for not picking out what I did wrong but...

500 N are acting 140 degrees, 150 N are acting at 90 degrees and 420 N are acting at 60 degrees. Find the resultant.

the Resultant = (Rx^2 + Ry^2)^(1/2)

The angle of the resultant = arctan(Fy/Fx)

Correct?

Well, Fx = -500cos40 + 150cos90 + 420cos60 = -173.02222

Fy = 500sin40 + 150sin90 + 420sin60 = 835.12447.

Resultant = ((8.35.12447)^2 + (-173.02222)^2)^(1/2) = 852.85964 N

angle of resultant = arctan(Fy/Fx) = arctan(835.12447/-173.02222) =
78.29.

The answers in the back of the book say 853 N and 101.7 degrees.

I have the resultant correct but not the angle. If I had both wrong, I wouldn't question it, but one right and one wrong makes me question it. What went wrong?
 
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  • #2
Originally posted by Sting
I should slap myself for not picking out what I did wrong but...

500 N are acting 140 degrees, 150 N are acting at 90 degrees and 420 N are acting at 60 degrees. Find the resultant.

the Resultant = (Rx^2 + Ry^2)^(1/2)

The angle of the resultant = arctan(Fy/Fx)

Correct?

Well, Fx = -500cos40 + 150cos90 + 420cos60 = -173.02222

Fy = 500sin40 + 150sin90 + 420sin60 = 835.12447.

Resultant = ((8.35.12447)^2 + (-173.02222)^2)^(1/2) = 852.85964 N

angle of resultant = arctan(Fy/Fx) = arctan(835.12447/-173.02222) =
78.29.

The answers in the back of the book say 853 N and 101.7 degrees.

I have the resultant correct but not the angle. If I had both wrong, I wouldn't question it, but one right and one wrong makes me question it. What went wrong?

Nothing, you are right. The angle you calculated is with respect to the resultant, which has an x-component along the negative x-axis. So, your resultant angle is 180 - 78.29 = 101.71 (since we take the angles starting along the positive x-axis)
 
  • #3
Okay, that's good that I'm not going crazy but...

my calculator needs to be changed because I got a slightly higher value for for the resultant angle.
 
  • #4
You were in the second quadrant.

atan is only defined for the first and fourth quadrants. You can either do a quadrant check or see if there is a function to take it into account (like atan2 in Matlab).
 
  • #5
Yes, but I still need either a new calculator or a new light bulb (I'm sure I misread the figure in the first place)
 
  • #6
There's a little screw in the back to adjust answers slightly!:smile:
 
  • #7
There's a little screw in the back to adjust answers slightly!

lol. It's really a good calculator but then I'm thinking about the conditions I worked that problem in. It was 3 in the morning, dark except for that faint blue-glow from my office lamp, and freaking cold (which I had to admit, I enjoyed). You know how solar powered calculators are in faint light...

So I'm assuming it was a midread figure due to bad lighting.

About that screw, do TI-89's have it? If there's any calculator I want to trample, it's that one. :smile:
 
  • #8
AAH Blasphemy!

I love my TI-89! You can teach them to do anything.

Mine came able to sit up, roll over, and fetch, and I've even taught it to make a mean cup of coffee.
 
  • #9
Blasphemy? Blasphemy is when you talk bad about Led Zeppelin.

But what I said about the TI-89... I meant every word of it! :smile:

Actually, I haven't gone beyond simple operations and simple graphing so maybe I'm just ignorant of its potential.

Would you mind sharing with us a few good (and useful) things you've acquired from you TI-89?
 
  • #10
Considering all the hours I spent calculating eigenvalues BY HAND, the TI-89 is a miracle!

(For doing basic things like evaluating functions or solving equations, a more basic calculator is easier to use.)
 
  • #11
Considering all the hours I spent calculating eigenvalues BY HAND, the TI-89 is a miracle!

I have to admit, they can take some of the grunt out of grunt work but I've wanted to smash mine into [al]0 pieces because the time I actually need it to work, it does nothing.

(For doing basic things like evaluating functions or solving equations, a more basic calculator is easier to use.)

You might think I'm weird but I've put away the calculator and I prefer to do most calculations by hand or mind.

I'm not trying to sound arrogant but I want to exercise the mind God blessed me with.

But don't get me wrong. If I have to calculate arctan(0.314573) in a hurry, forget Taylor Series. I'll use the calculator. :smile:
 
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FAQ: What went wrong with finding the resultant?

1) What is a vector resultant problem?

A vector resultant problem involves finding the magnitude and direction of the resultant vector when two or more vectors are added together.

2) How do you solve a vector resultant problem?

To solve a vector resultant problem, you must first determine the individual components (magnitude and direction) of each vector. Then, add the components of each vector together to find the resultant vector. Finally, use trigonometric functions to find the magnitude and direction of the resultant vector.

3) Can you use the Pythagorean theorem to solve a vector resultant problem?

Yes, the Pythagorean theorem can be used to solve a vector resultant problem if the vectors are at right angles to each other. In this case, the resultant vector can be found by using the equation c = √(a^2 + b^2), where a and b are the individual components of the vectors and c is the magnitude of the resultant vector.

4) What is the difference between a scalar and a vector quantity?

A scalar quantity is a physical quantity that only has magnitude, such as speed or temperature. A vector quantity has both magnitude and direction, such as velocity or force.

5) Are there any real-world applications of vector resultant problems?

Yes, vector resultant problems are commonly used in physics and engineering to calculate the net force or velocity of an object. They are also used in navigation and mapping to determine the direction and distance between two points.

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