Friction on Aluminum & Copper Blocks: Find Sum of Magnitudes

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The discussion centers on a physics problem involving a 2.00-kg aluminum block and a 6.00-kg copper block connected by a string over a frictionless pulley on a steel surface at a 30-degree incline. It is determined that the system will not start moving because the force attempting to move the copper block down the incline (29.4 N) is less than the total frictional force acting on both blocks (38.9 N). Participants clarify the calculations for static friction and the normal force, emphasizing the importance of using the correct coefficients of friction. There is a focus on ensuring the correct application of the friction formula, which involves the coefficient of friction multiplied by the normal contact force. The conversation highlights the need for accurate understanding and application of physics principles in solving the problem.
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Homework Statement


A 2.00-kg aluminum block and a 6.00-kg copper block are connected by a light string over a frictionless pulley. They sit on a steel surface as shown in Figure P5.64, where theta= 30.0°. When they are released from rest, will they start to move? If so, determine (a) their acceleration and (b) the tension in the string. If not, determine the sum of the magnitudes of the forces of friction acting on the blocks.

https://drive.google.com/file/d/0B4ksjSfetGUOeUdXX0tKODRWVjg/view?usp=sharing ---- IMAGE
fyzika.png


Homework Equations


1) I don't know how to calculate the last part: "If not, determine the sum of the magnitudes of the forces of friction acting on the blocks."

2) Is it so far correctly calculated?

The Attempt at a Solution


For the system to start to move when released, the force tending to move m2 down the incline, m g 2 sinθ, must exceed the maximum friction force which can retard the motion:

F = f(steel) + f(copper)

= µ1m1g + µ2m2g

= 0.61 x 2.0 x 9.8 + 0.53 x 6.0 cos(30) x 9.8 = 38.9 N

Force tending to cause the system to move = 6.0 x 9.8 x sin(30) = 29.4 N
Hence it will not start to move when released.
 
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It just hit me... Did I already solved the: "If not, determine the sum of the magnitudes of the forces of friction acting on the blocks." Part? :D
 
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Ondra said:
µ2m2g
That is not correct for the block on the slope, but I see you used the right version in the next line.
Ondra said:
Did I already solved the: "If not, determine the sum of the magnitudes of the forces of friction acting on the blocks." Part?
Possibly. You calculated two forces. Which one are you suggesting as answer to part b?
Before answering, what exactly is the standard equation relating to static friction?
 
The formula for friction is F = coeffiicent of friction * Normal contact force. For M2, the normal contact force is not the weight. but you have corrected it lol. Where did you get the coefficients from? Online? I see its 0.61 between aluminium and steel.

Also, I know I posted in your earlier thread that got deleted, so please try to check out these two websites:

scroll down for mechanics:
http://www.cimt.org.uk/projects/mepres/alevel/alevel.htm
 
Bipolar Demon said:
The formula for friction is F = coeffiicent of friction * Normal contact force.
Not quite, and that is important here.
 
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