Resultant Force: Something I am messing up on in calculation - ?

AI Thread Summary
The discussion revolves around a resultant force calculation where the original poster is confused about their answer compared to the correct one. They used a TI-89 Titanium calculator but ended up with a significantly different result. After some troubleshooting, it was determined that the calculator was set to the wrong mode, which affected the calculations. Switching to a simpler Casio calculator resolved the issue, leading to the correct answer of 3559.2 Newtons. The conversation highlights the importance of ensuring the calculator is in the correct mode for accurate results.
nukeman
Messages
651
Reaction score
0
Resultant Force: Something I am messing up on in calculation - ?

Homework Statement



Hey all,

This is a resultant force word problem. I will just post the my answer and the problems answer. I have no idea why there is a different answer here.

Correct: R^2=1500^2+2300^2 - 2 x 1500 x 2300 Cos(138)
Correct Answer = R = 3559.2 Newton

My answer is WAYYY off, so there is something I am missing here...

By the way, I am using the TI-89 Titanium

Thanks for any help!


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org


nukeman said:

Homework Statement



Hey all,

This is a resultant force word problem. I will just post the my answer and the problems answer. I have no idea why there is a different answer here.

Correct: R^2=1500^2+2300^2 - 2 x 1500 x 2300 Cos(138)
Correct Answer = R = 3559.2 Newton

My answer is WAYYY off, so there is something I am missing here...

By the way, I am using the TI-89 Titanium

Thanks for any help!


Homework Equations





The Attempt at a Solution


I get the correct answer just typing it into a cheap ativa scientific calculator.

Can you say what keystrokes you used to get your incorrect answer? I did something like:

cos 138 =

* 2300 =

* 1500 =

* -2 =

+ 2300^2 =

+ 1500^2 =

SQRT =
 


Maybe it's TOO much calculator for too SMALL a problem.

First question: Is you calculator in 'degree' mode, like berkeman's obviously is?
 


Thanks both of you.

I figured it out. Sammy, you were basically right, it was too much of a calc for it. Yes, I was in the wrong mode. PAIN in the butt!

So, pulled out my dusty Casio calculator and BOOM! solved :)

Thanks guys
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top