# Resultant Force on an object during Non-Uniform Motion

1. Oct 31, 2012

### mldavis086

1. The problem statement, all variables and given/known data

A 0.30-kg mass attached to the end of a string swings in a vertical circle
(R = 1.6 m), as shown. At an instant when θ = 50°, the tension in the string is
8.0 N. What is the magnitude of the resultant force on the mass at this instant?

http://www.flickr.com/photos/89533422@N08/8142258633/in/photostream/lightbox/

2. Relevant equations

m*centripetal acceleration=m*v^2/r
mg
basic trig

3. The attempt at a solution

I've figured out
mg = 2.94 N
mg sin 50 = 2.25 N (which acts opposite of the direction of the mass)
mg cos 50 =1.89 N (the tension due to gravity)
8-1.89 = 6.11 N (the tension due to motion of the mass)

8=0.3*v^2/1.6 (v=5.71 m/s)
-2.25=0.3*a (a=-7.5 m/s^2)

I'm not 100% on the last 2

But I still don't know what the resultant force is? Or even what force they are referring to. If anyone can help it would be greatly appreciated!

2. Oct 31, 2012

### SammyS

Staff Emeritus

What are all of the forces acting on the mass ?

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3. Oct 31, 2012

### mldavis086

The only force acting on it is gravity right? Does that make the 'resultant force' the 2.25N opposite of it's direction at the moment?

4. Oct 31, 2012

### mldavis086

Or is it just mg=2.94N? I am confused with the term 'resultant force' I think

5. Oct 31, 2012

### mldavis086

Wait the centripetal acceleration is a force too right? 8N towards the center of the circle.

6. Oct 31, 2012

### mldavis086

I think I get it. The 2 forces are the force towards the center, and the force of gravity straight down. If I calculate the 'resultant' of these 2 vectors. I get a force of 2.2N up and 6.13N left. Then using Pythagoreans Theorem. The magnitude of the resulting vector is 6.51N. Can anyone out there confirm if I am looking at this problem properly? I really want to make sure I understand. Thanks

Last edited: Oct 31, 2012
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