Two blocks attached by a string(dynamic)

  • Thread starter Thread starter MikeMai
  • Start date Start date
  • Tags Tags
    Blocks Two blocks
Click For Summary
The problem involves two blocks connected by a cord, with a force applied to one block on a horizontal surface. The user attempts to find the tension in the rope using Newton's second law, accounting for friction. They derive two equations based on the net forces acting on each block, incorporating the coefficient of friction. After substituting and solving the equations, they calculate the tension to be approximately 35N. The solution is confirmed as correct by another forum member.
MikeMai
Messages
5
Reaction score
0
1. Homework Statement
Two blocks (m1=4.0kg, m2=7.5kg) are attached by a light cord pulled along a horizontal surface by a force of 53N as shown in the diagram. If the coefficient of friction between the blocks and the surface is 0.30, what is the tension in the rope between the two blocks.

M2 -------- M1 ------->Force applied


2. Homework Equations
Fnet= Ma


3. The Attempt at a Solution

I used the Fnet=ma, on m1 i got Fapplied - T - Ff = ma
But i don't know how to find the acceleration. And i know there's two different frictions because of two different mass. So i don't know how to join them together and find Tension.

ps: I just joined this forum, What should i do if i have a a lot of questions or simply want someone to check over my assignment for me before I hand them in? (got like 20-30questions). I don't want to violate the rule so I am just wondering if there's a specific section to post or something.
 
Physics news on Phys.org
What about M2?
Apply Newton2 to M2 and then you'll have 2 equations to work with.

If you have other questions you want to post, then create a new thread for each one.
 
Hmm, ok so now i got this.

m1:
Fnet=m1a
Fapplied-T-Ff=m1a

m2:
Fnet=m2a
T-Ff=m2a
a=T-Ff/m2

plug a into m1's equation

Fapplied-T-Ff=m1(T0Ff/m2)
53-T-(0.30x4.0x9.8)=4.0(T-0.30x7.5x9.8/7.5)
53=1.53T
T=34.565N = 35N

If someone can check if i did the calculation right, it would be greatly appreciated.

Thank you.
 
Yes, this is correct :)
 
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
21
Views
825
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 45 ·
2
Replies
45
Views
4K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K