What is the tension in the string between two blocks with friction?

In summary, the online program disagrees with the answer that T = 10.04 N. The two tensions in the string are different, which is why the online program gives a different answer.
  • #1
Mivz18
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!
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  • #2
You could draw an FBD for both of the blocks with separate tension values for both of the sections of string. Then you should be able to set up, and solve, a system of equations.
 
  • #3
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?
 
  • #4
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??
 
  • #5
Start by indentifying all the horizontal forces on m1. (The tension in the string is not the only force.) Apply Newton's 2nd law.
 
  • #6
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?
 
  • #7
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!
 
  • #8
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.
 
  • #9
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.)
 

1. What is the concept of "Two Blocks with friction"?

The concept of "Two Blocks with friction" is a physics problem where two blocks are in contact with each other and are moving on a surface with friction. The problem involves calculating the acceleration and motion of the blocks based on their masses, the coefficient of friction between them and the surface, and the external forces acting on them.

2. How do you calculate the acceleration of the blocks in "Two Blocks with friction"?

To calculate the acceleration of the blocks, you need to use Newton's Second Law which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma). You also need to take into account the frictional force between the blocks and the surface, which is calculated using the coefficient of friction and the normal force between the blocks and the surface.

3. What factors affect the motion of the blocks in "Two Blocks with friction"?

The motion of the blocks is affected by several factors, including their masses, the coefficient of friction between them and the surface, the external forces acting on them, and the surface itself. Additionally, the angle of the surface and the presence of any other objects or obstacles can also affect the motion of the blocks.

4. How do you determine the direction of motion for each block in "Two Blocks with friction"?

The direction of motion for each block can be determined by analyzing the forces acting on the blocks. If the net force on a block is in the same direction as its initial velocity, the block will continue to move in that direction. However, if the net force is in the opposite direction, the block's motion will be slowed down or it may even change direction.

5. What real-life situations can be modeled using the concept of "Two Blocks with friction"?

The concept of "Two Blocks with friction" can be applied to various real-life situations, such as a car moving on a road with friction, a person pushing or pulling an object on the ground, or two objects sliding against each other. It can also be used to understand the motion of objects on an inclined plane or on a rough surface.

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