Newton's Second Law (Pulley Sustem)

Click For Summary
The discussion focuses on a physics problem involving two masses, M1 (4.0 Kg) and M2 (20 Kg), connected by a rope over a frictionless pulley, with M1 sliding horizontally and experiencing kinetic friction. The coefficient of kinetic friction is given as uk = 0.638. Participants are advised to draw free body diagrams (FBD) for each mass and apply Newton's Second Law to derive equations for the system. The goal is to calculate the acceleration of the masses and the tension in the rope. Additionally, tips for posting equations using LaTeX and image uploading are provided for clarity in future discussions.
armn91
Messages
2
Reaction score
1
Homework Statement
I am trying to find the solution to the following problem. Could you please let me know if what I have done is correct? Thank you.
Relevant Equations
Newton's Second Law
Two masses, M1=4.0 Kg, and M2=20 Kg, are attached by a rope that crosses over a massless, frictionless pulley. M1 slides horizontally and experiences a kinetic frictional force due to its motion across the surface. The coefficient of kinetic friction is uk = 0.638. When solving the problem, start drawing a FBD for each mass and writing appropriate N1L and N2L equations for each mass. Solve the system of equations for find the acceleration of the masses and the tension of the rope (see figure in document attached).
 

Attachments

  • IMG_0288.jpg
    IMG_0288.jpg
    50.3 KB · Views: 118
  • IMG_0289.jpg
    IMG_0289.jpg
    22.4 KB · Views: 126
Physics news on Phys.org
It looks correct to me. For future reference: After you upload a figure click "Insert" + Full image. Full images are easier to read than thumbnails. Also, for "work shown" we prefer equations in LaTeX to photos of handwritten notes. Click "LaTeX Guide", lower left, to learn how to do it.
 
  • Like
Likes berkeman and MatinSAR
Ok, thank you very much for your help!
 
Welcome to PF.

I will send you some tips for posting equations using LaTeX. :smile:
 
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 22 ·
Replies
22
Views
6K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
4
Views
792
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 21 ·
Replies
21
Views
10K
  • · Replies 18 ·
Replies
18
Views
1K