Resultant Force on an object during Non-Uniform Motion

mldavis086
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Homework Statement



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/

Homework Equations



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

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!
 
on Phys.org
mldavis086 said:

Homework Statement



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/

Homework Equations



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

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!

attachment.php?attachmentid=52521&stc=1&d=1351730446.jpg


What are all of the forces acting on the mass ?
 

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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?
 
Or is it just mg=2.94N? I am confused with the term 'resultant force' I think
 
Wait the centripetal acceleration is a force too right? 8N towards the center of the circle.
 
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:

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