Friction force of 2 connected block on steel surface

[EnricoHendro]

Homework Statement

An aluminum block of
mass m1 = 2.00 kg and a copper block of mass m2 = 6.00 kg are connected by a light string over a frictionless pulley. They sit on a steel surface as shown in Figure P5.84, where θ = 30.0°. (a) When they are released from rest, will they start to move? If they do, determine (b) their acceleration and (c) the tension in the string. If they do not move, determine (d) the sum of the
magnitudes of the forces of friction acting on the blocks.

Relevant Equations

fs= cofficient of static friction x n
where : n = m2.gcosθ (for copper)
n = m1.g (for alumunium)
F = m2.gsinθ - T - fs (for Copper)
F = T - fs (for Alumunium)

hello there, I am wondering if my attempt at a solution for this particular problem correct?? I mean I know that the blocks won't move and the total friction force is 38.95N (at least lesser than that, since that is the maximum static friction). I saw someone attempted the solution but it is completely different than mine (although we arrive at the same conclusion). Thank you

 

[EnricoHendro]

I had a little miscalculation at m2.gsinθ...It should be 29.4 N (which is still less than 38.95N)

 

[haruspex]

I'm confused about the friction coefficient for steel/ aluminium. You seem to have stated it as 0.53 (but I cannot read the suffix), in the calculations I see 0.61, and online it says 0.47.

 

[EnricoHendro]

I'm confused about the friction coefficient for steel/ aluminium. You seem to have stated it as 0.53 (but I cannot read the suffix), in the calculations I see 0.61, and online it says 0.47.

the 0.53 is the coefficient of steel/ copper. I wrote the coefficient for steel/ alumunium to be 0.61 (I wrote it below the "for alumunium" at the right side, below the copper diagram (I forgot to write it down, so I kinda add it beside the "for copper". Thus in the calculation you see 0.61
p.s. I am using the friction coefficient from my textbook

 

[haruspex]

the 0.53 is the coefficient of steel/ copper. I wrote the coefficient for steel/ alumunium to be 0.61 (I wrote it below the "for alumunium" at the right side, below the copper diagram (I forgot to write it down, so I kinda add it beside the "for copper". Thus in the calculation you see 0.61
p.s. I am using the friction coefficient from my textbook

In that case I agree with 39N for the sum of magnitudes of the two maximum frictional forces. (Shouldn’t quote more sig figs than that since the coefficients are only given to two.)

 

[EnricoHendro]

In that case I agree with 39N for the sum of magnitudes of the two maximum frictional forces. (Shouldn’t quote more sig figs than that since the coefficients are only given to two.)

I see...so what about my method/analysis for that particular problem?? do you think my analysis is correct?? I'm really concern about the analysis since my other objective of studying physics to improve/test my analytical/logical thinking :)

 

[haruspex]

I see...so what about my method/analysis for that particular problem?? do you think my analysis is correct?? I'm really concern about the analysis since my other objective of studying physics to improve/test my analytical/logical thinking :)

Yes, it was a reasonable way to find the max total friction force (as a sum of magnitudes, which is what you need). But you do understand that you have not answered part d, right?

 

[EnricoHendro]

Yes, it was a reasonable way to find the max total friction force (as a sum of magnitudes, which is what you need). But you do understand that you have not answered part d, right?

oh yeah, I totally forgot about part d...thanks for reminding though