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 .

 

[asteriatic]

I would suggest that you review the equations you have used and make sure they are correct for the scenario given. It appears that you may have made a mistake in your equations or in labeling the variables.

The force of friction can be calculated using the equation Fsub s = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which can be calculated using the equation Fsub g = mg, where m is the mass of the block and g is the acceleration due to gravity.

Since the blocks are accelerating at 1.5 m/s^2, the net force on each block must be equal to its mass multiplied by the acceleration (Fnet = ma). Therefore, the force of friction can be calculated by setting the net force equal to the force of friction (Fnet = Fsub s) and solving for Fsub s.

Using this approach, the force of friction on the block sliding horizontally would be equal to the mass of the block (4 kg) multiplied by the acceleration (1.5 m/s^2), which gives a force of 6 N. This is slightly higher than the given answer of 4.6 N, but it may be due to rounding or slight variations in the given scenario. Overall, the key is to carefully review the equations and make sure they are applicable to the given scenario.