How Do You Calculate the Tension T2 Between Blocks in a Frictional System?

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Homework Statement


There are three blocks on a table. They are connected by a massless cord and pulled to the right. From rigt to left, it follows: T3, block 3, T2, block 2, T1, block 1
m1: 2.5kg
m2: 9.5kg
m3: 1.5kg
mu kinetic: .051
The pulling force is equal to T_3 = 98.0 N. What is the tension T_2?

Homework Equations


I think my equations may be incorrect, but here goes
F[tex]_{}fr[/tex]=F[tex]_{}N[/tex] + [tex]\mu_{}k[/tex]
F[tex]_{}pull[/tex]=F[tex]_{}fr[/tex] + ma

The Attempt at a Solution



I solved for accel. and then plugged it into the equation for tension and I got a value of 81.11 N, but that is not correct.

Any help please?
 
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BeatTheRuckus said:

Homework Statement


There are three blocks on a table. They are connected by a massless cord and pulled to the right. From rigt to left, it follows: T3, block 3, T2, block 2, T1, block 1
m1: 2.5kg
m2: 9.5kg
m3: 1.5kg
mu kinetic: .051
The pulling force is equal to T_3 = 98.0 N. What is the tension T_2?

Homework Equations


I think my equations may be incorrect, but here goes
F[tex]_{}fr[/tex]=F[tex]_{}N[/tex] + [tex]\mu_{}k[/tex]
F[tex]_{}pull[/tex]=F[tex]_{}fr[/tex] + ma

The Attempt at a Solution



I solved for accel. and then plugged it into the equation for tension and I got a value of 81.11 N, but that is not correct.

Any help please?
[/b]
The friction force is not correct. For objects in motion, it's [tex]F_{friction} = \mu_k(N)[/tex], where N is the normal force between the block and table. You're going to have to calculate it for each block, then use Newton 2 and 'free body diagrams'. Are you familiar with them?