Resultant Force on an object during Non-Uniform Motion

Click For Summary
SUMMARY

The discussion centers on calculating the resultant force acting on a 0.30-kg mass swinging in a vertical circle with a radius of 1.6 m at an angle of 50°. The tension in the string is given as 8.0 N. The gravitational force (mg) is calculated to be 2.94 N, with components mg sin 50° = 2.25 N and mg cos 50° = 1.89 N. The resultant force is determined by combining the tension and gravitational forces, leading to a final magnitude of approximately 6.51 N using vector addition.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula (m*v^2/r)
  • Basic knowledge of trigonometric functions (sine and cosine)
  • Familiarity with Newton's laws of motion
  • Ability to perform vector addition and apply the Pythagorean theorem
NEXT STEPS
  • Study the concept of resultant forces in circular motion
  • Learn how to apply Newton's second law to non-uniform motion
  • Explore the effects of tension in circular motion scenarios
  • Review vector decomposition and addition techniques in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for examples of resultant force calculations in real-world applications.

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: 1,126
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:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 7 ·
Replies
7
Views
9K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K