• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Two Blocks with friction

  • Thread starter Mivz18
  • Start date
44
0
Ok, I have this problem I've been working on forever, but can't figure out the last part of the question.

A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of P = 30 N. The coefficient of friction of the blocks with the floor is µ = 0.1.

First it asked what the acceleration of the two blocks were, and I figured that out to be 5.02 m/s ^2 . Then the second part asks what is the tension of the string between the two blocks. This is where my trouble begins.

I know that [/SIZE]Fnet = T on m1, unless I'm mistakenly not counting the friction. However, with the force of 30 N to the right and 4.9 N of friction to the left in the system, that is not accounted for in the Fnet of 10.04 N for m1 and 15.06 for m2 that I obtained. If T is the only force on m1, wouldn't T = 10.04 N ?? Please help!
PHP:
[CODE]
 

Attachments

Last edited:

NateTG

Science Advisor
Homework Helper
2,449
5
You could draw an FBD for both of the blocks with seperate tension values for both of the sections of string. Then you should be able to set up, and solve, a system of equations.
 
575
2
You know the acceleration and mass of each block. You also know the applied force and coeff. of friction. Since [itex]F_{net}=ma[/itex], can you develop the equation to include [itex]F_{tension}[/itex], which you can then isolate to answer the question?
 
44
0
I had tried developing an equation but it doesn't seem to work like I had stated. Since in m1 Fnet = T, then T should = 10.04 N ? The online program I am using doesn't agree with that answer, so I'm kind of stumped here. I have drawn FBD for both blocks but haven't been able to figure it out. HELP!?!?
 

Doc Al

Mentor
44,795
1,051
Start by indentifying all the horizontal forces on m1. (The tension in the string is not the only force.) Apply Newton's 2nd law.
 
44
0
Ok, maybe I'm missing something but this is what I get:

on m1, of 2 kg, you obtain two forces, that of friction to the left, and that of tension to the right. Therefore, Fnet = T - friction, or T- friction = ma . From the acceleration I obtained in the first part, 5.02 , T - friction = 10.04.

on m2, of 3 kg, I obtain three forces, that of the applied 30 N to the right and that of friction and tension to the left. Fnet = 30 -T - friction , or 30 - T -friction = ma . From the acceleration of 5.02 i get that T + friction = 14.94 .

Then I add the two equations to solve for the variable, Tension, that I'm looking for.

T - friction = 10.04
T + friction = 14.94

2T = 24.98
T = 12.49

However, the online program disagrees with that answer. Did I calculate something wrong, or am I assuming that the two T's are equal and they really aren't?
 

Doc Al

Mentor
44,795
1,051
Mivz18 said:
on m1, of 2 kg, you obtain two forces, that of friction to the left, and that of tension to the right. Therefore, Fnet = T - friction, or T- friction = ma . From the acceleration I obtained in the first part, 5.02 , T - friction = 10.04.
Right, but "friction" is not an unknown! Figure it out. Then solve for T. And you're done. :smile:

on m2, of 3 kg, I obtain three forces, that of the applied 30 N to the right and that of friction and tension to the left. Fnet = 30 -T - friction , or 30 - T -friction = ma . From the acceleration of 5.02 i get that T + friction = 14.94 .
Right, but not needed. Once again "friction" can be figured out.

Then I add the two equations to solve for the variable, Tension, that I'm looking for.

T - friction = 10.04
T + friction = 14.94
Big mistake! The two things you label "friction" are not the same!
 
44
0
nevermind, lol, I found my mistake. I was in the wrong in adding the equations. Instead,

T - friction = 10.04
T = 10.04 + friction = 10.04 + 1.96 = 11.98

T + friction = 14.94
T = 14.94 - friction = 14.94 - 2.94 = 12

So basically, both are about 12 N , and the online program accepted it. Thanks everyone for your guidance.
 

Doc Al

Mentor
44,795
1,051
tension is the same throughout the string

Mivz18 said:
So basically, both are about 12 N , and the online program accepted it.
Just to be clear: There is only one tension in the string. No need to solve for it twice! (Any difference in the two answers is just due to rounding errors.)
 

Related Threads for: Two Blocks with friction

Replies
7
Views
8K
  • Posted
Replies
1
Views
3K
  • Posted
Replies
1
Views
4K
Replies
1
Views
3K
Replies
5
Views
5K
Replies
2
Views
2K
Replies
5
Views
298
Replies
3
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top