Resultant Force on an object during Non-Uniform Motion

AI Thread Summary
The discussion revolves around calculating the resultant force on a 0.30-kg mass swinging in a vertical circle at an angle of 50°. The tension in the string is given as 8.0 N, while the gravitational force acting on the mass is calculated to be 2.94 N. The user explores the components of gravitational force, finding that the vertical component (mg sin 50) is 2.25 N and the horizontal component (mg cos 50) is 1.89 N. After considering both the tension and gravitational forces, the user attempts to calculate the resultant force using vector addition, arriving at a magnitude of approximately 6.51 N. The discussion emphasizes understanding the concept of resultant force as the combination of gravitational and centripetal forces acting on the mass.
mldavis086
Messages
10
Reaction score
0

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!
 
Physics news 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 ?
 

Attachments

  • 8142258633_7f77e95fba_s.jpg
    8142258633_7f77e95fba_s.jpg
    1.5 KB · Views: 1,118
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:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top