Friction with two blocks and tension

  • #1
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Homework Statement


A 2.00kg aluminium block and a 6.00kg copper block are connected by a light string over a frictionless pulley. They sit on a steel surface, which is fixed, as shown in the figure, with a (let it be alpha lol) = 43.0deg. The coefficient of static friction for aluminum on steel is 0.56, and for copper on steel is 0.83. Determine whether the blocks start to move once any holding mechanism is released. Then, calculate the sum of the magnitudes of the forces of friction acting on the blocks.

Homework Equations


F = ma


The Attempt at a Solution


The blocks should not move at all according to this equation i've derived (i'm not sure if i should use static or kinetic friction for this case).

a = (m1gsin(a)-us1m1gcos(a)-us2m2g)/(m1+m2)

Where us1 is the coefficient of static friction of copper, m1 is the mass of copper and us2 is the coefficient of static friction of aluminum.

This gives us a negative acceleration and thus the system won't move.

I've determined that the forces of friction for each of the two blocks are the following:

fr1 = us1m1gcos(a) = 10.9872N
fr2 = us2m2g = 35.7294N

I'm using the coefficient of static friction as the two blocks don't move (again not too sure on this one). I've added these two frictions and i got that the final result (46.7N) was incorrect. Or should i do vector addition and determine the magnitude? However, i also did the same and got a wrong result (44.4N).

What should i do?

By the way: i have the following FBDs:

For Al, i have the normal force going upwards, the weigh, the friction and the tension. For Cu, i have the tension going up the plane, the friction up the plane, the normal force perpendicular to the surface and the weigh.
 

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  • #2
Simon Bridge
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Homework Statement


A 2.00kg aluminium block and a 6.00kg copper block are connected by a light string over a frictionless pulley. They sit on a steel surface, which is fixed, as shown in the figure, with a (let it be alpha lol) = 43.0deg. The coefficient of static friction for aluminum on steel is 0.56, and for copper on steel is 0.83. Determine whether the blocks start to move once any holding mechanism is released. Then, calculate the sum of the magnitudes of the forces of friction acting on the blocks.

Homework Equations


F = ma


The Attempt at a Solution


The blocks should not move at all according to this equation i've derived (i'm not sure if i should use static or kinetic friction for this case).
Consider what the words "static" and "kinetic" mean in relation to the words "moving" and "not moving". Is "static" the sort of word you normally associate with movement or with non-movement?

a = (m1gsin(a)-us1m1gcos(a)-us2m2g)/(m1+m2)

Where us1 is the coefficient of static friction of copper, m1 is the mass of copper and us2 is the coefficient of static friction of aluminum.
So m1 is copper and m2 is aluminium ... let me tidy that up for you:

$$a=m_{Cu}g\sin\alpha - (\mu_{Cu}m_{Cu}+\mu_{Al}m_{Al})g\cos\alpha$$

This gives us a negative acceleration and thus the system won't move.
- negative acceleration could be an increase in speed the other way from positive - I think you need to say more than that.

What you mean is that the force supplied by gravity is smaller than the maximum that the static friction can provide.

I've determined that the forces of friction for each of the two blocks are the following:

fr1 = us1m1gcos(a) = 10.9872N
fr2 = us2m2g = 35.7294N
Those are the maximum forces that the friction could provide.
However, the actual friction does not need to be so high.
 
  • #3
haruspex
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$$a=m_{Cu}g\sin\alpha - (\mu_{Cu}m_{Cu}+\mu_{Al}m_{Al})g\cos\alpha$$
I think you meant
$$a(m_{Cu}+m_{Al})=m_{Cu}g\sin\alpha - \mu_{Cu}m_{Cu}g\cos\alpha-\mu_{Al}m_{Al}g$$
 
  • #4
haruspex
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I've determined that the forces of friction for each of the two blocks are the following:

fr1 = us1m1gcos(a) = 10.9872N
fr2 = us2m2g = 35.7294N
I think you've swapped over the two blocks. m2 is the 6kg copper block on the slope.
 
  • #5
Simon Bridge
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I think you meant
$$a(m_{Cu}+m_{Al})=m_{Cu}g\sin\alpha - \mu_{Cu}m_{Cu}g\cos\alpha-\mu_{Al}m_{Al}g$$
... actually I meant only to transliterate OPs equation quoted in the same post. ;)

Still - I should have picked that up.
 
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  • #6
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I'm so sorry for not replying! (midterms are killing me!)

I figured out the problem. I used the static friction of the blocks and then i determined that the acceleration was negative (thus the blocks aren't moving as the treshold friction isn't surpassed). To obtain the final friction force, i just added up the termins in the equation. We have that mCugsinα = T + fr1 (however, remember that T is actually fr2!) So you just add them up and get the final result.

Thank you Simon.
 

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