How Is the Force of Friction Calculated for a Block Sliding Horizontally?

[yosimba2000]

Homework Statement

Three blocks are released from rest and accelerated at 1.5 m/s^2. What is the magnitude of the force of friction on the block sliding horizontally?

Homework Equations

F net = Mass * Acceleration
(adding up 3 EQ's and canceling out variables)

*Tension # = Tension sub #
* Force of gravity = F sub g
* Force of friction = F sub s

The Attempt at a Solution

I made three equations: F sub g - T sub 1 = 4a <---- Block on farthest right
T sub 1 - F sub s - T sub 2 = 4a <---- block on table
T sub 3 - F sub g = 2a <--- farthest left block

*after canceling out variables from combing the EQ's*

4a+4a+2a = F sub s - T sub 2 + T sub 3

I do not know how to find the force of friction because i cannot find the tensions.

I used F sub 1 two times because i thought since the 4kg boxes are equal in mass and are accelerating at the same value, the tension must be equal.

The answer is 4.6 N. I think I labeled some tension wrong I think.

Thanks!

 

[SammyS]

Isn't it true that T₃ = T₂ ?

 

[yosimba2000]

if T2 = T3, doesn't that mean they will cancel out after combining the 3 EQ's. giving 10a = -(force of friction)? and then after solving, you get 15 N. but the book says 4.6 N

 

[SammyS]

The middle mass (the one on top) exerts a force on the left hand mass, that's equal & opposite to the force the mass on the left exerts on the middle mass.

∴, T₂ = T₃ .