Blocks, pulleys, more friction woes

  • Thread starter 1MileCrash
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  • #1
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Homework Statement



Body A weighs 100 N, and body B weighs 35 N. The coefficients of friction between A and the incline are μs = 0.56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x axis be up the incline.

06_38.gif


Find acceleration along x.

Homework Equations





The Attempt at a Solution



Force of smaller block is just is weight, since it's just freely hanging.

Force of larger block is lessened because it is on an incline and due to friction. This means it will be pulled up the incline?

35 - u100cos40 = ma
15.85 = ma
6.3096 = a

Is my head on right?
 

Answers and Replies

  • #2
PeterO
Homework Helper
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Homework Statement



Body A weighs 100 N, and body B weighs 35 N. The coefficients of friction between A and the incline are μs = 0.56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x axis be up the incline.

06_38.gif


Find acceleration along x.

Homework Equations





The Attempt at a Solution



Force of smaller block is just is weight, since it's just freely hanging.

Force of larger block is lessened because it is on an incline and due to friction. This means it will be pulled up the incline?

35 - u100cos40 = ma
15.85 = ma
6.3096 = a

Is my head on right?
Firstly, the small block also has the string pulling up; it is not just the weight force that acts on it.

Which value did you use for μ ?

You should use μs at first, to check that the 35N weight is sufficient to start the large block at all - and don't forget the component of the 100N block's weight acting down the slope!! If not enough force to get things moving, the acceleration will be zero.
Be sure to consider the possibility tat the block might slide down the slope

If, however, the force is strong enough to get things moving, you then use μk to calculate the net force on the masses as they accelerate.

It looks like they are assuming g = 10, so the masses are 10kg and 3.5kg giving a total of 13.5 kg
 
  • #3
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I'm a bit confused about the force of the string. How do I find what the string is doing if it depends on the force exerted on both blocks, which depends on the force of the string?
 
  • #4
PeterO
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I'm a bit confused about the force of the string. How do I find what the string is doing if it depends on the force exerted on both blocks, which depends on the force of the string?
If you consider the two blocks and the string as a single system, then the tension in the string is an internal force so is not involved in the global calculation.
Newtons Laws refer to unbalanced external forces causing motion.
 
Last edited:
  • #5
PeterO
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I'm a bit confused about the force of the string. How do I find what the string is doing if it depends on the force exerted on both blocks, which depends on the force of the string?
Follow-up answer to previous post: read in conjunction
If the table was smooth, and horizontal, the only force acting in the direction the system might move is the weight of 35N
Given the mass of the system is 135/g , we can establish the resulting acceleration.
If we analyse the system, the tension in the string is all that accelerates the "100N" mass so we can use F = ma to find the size of tension.

In this problem, the table is not flat, so in addition to the 35N weight force trying to accelerate the system, there is a component of the 100N weight force trtying to get things going in the other direction. Which one wins??
Also there is the friction Force trying to stop - or at least slow down, the movement.

Another hint:
We know that the components of the 100N along the slope [trying to cause motion] and perpendicular to the slope [to be joined with the co-efficient of friction to give the maximum frictional force] are 100sin40 and 100cos40; but which one is which.

Well if the angle was much smaller, the Force along the slope would be practically zero. is it 100sin(x) or 100cos(x) that is almost zero when x is very small - that way you can show yourself which one is correct rather than trying to remember which one is correct.
 
Last edited:

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