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Calculate Tension in the String

  • #1
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Homework Statement


Two blocks are connected by a massless string that passes over a frictionless pulley. The coefficient of static friction between m1 and the table is 0.45. The coefficient of kinetic friction is 0.35. Mass m1 is 45kg and mass m2 is 12kg. Determine tension in the string. ***I attached a picture of the diagram provided with the question.

Homework Equations


Fnet=ma
μsFN=Fs

The Attempt at a Solution


I thought that (Fnet)x=0 for m1 since this system is not in motion- it's in static equilibrium. I worked with the forces acting on m1, assuming that Fnet=0 for m1.
So, (Fnet)x=Fs+Ft
0=μsFN+Ft
-(0.45)(mg)=FT
-(0.45)[(45)(9.80]=Ft
-198.45N=Ft

The answer in the back of the book is 120N. What did I do wrong?
IMG_6710.jpg
 
Last edited:

Answers and Replies

  • #2
CWatters
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Hint: The friction force is not 0.45*mg. That is the maximum possible friction force.
 
  • #3
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Hint: The friction force is not 0.45*mg. That is the maximum possible friction force.
So you calculate friction using μkFn?
 
  • #4
CWatters
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You are over thinking the problem. What if m1 was replaced by a nail hammered into the table. What would determine the tension?
 
  • #5
vela
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For static friction, you have ##F_s \le \mu_s F_n## (as CWatters mentioned above). If you don't understand the implication of this expression, consider the situation where there's no string connected to mass 1. It's just sitting on the surface. What is the force of static friction on it?
 
  • #6
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You are over thinking the problem. What if m1 was replaced by a nail hammered into the table. What would determine the tension?
The net force on the nail?
 
  • #7
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For static friction, you have ##F_s \le \mu_s F_n## (as CWatters mentioned above). If you don't understand the implication of this expression, consider the situation where there's no string connected to mass 1. It's just sitting on the surface. What is the force of static friction on it?
0N- there's no attempted motion
 
  • #8
CWatters
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Correct.
 
  • #9
CWatters
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The net force on the nail?
Try thinking about the other end of the rope!
 
  • #10
vela
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So ##F_s \ne \mu_s F_n## in that case since ##F_s = 0## but ##\mu_s F_n > 0##. The same is true in the original problem. You can't really say much about ##F_s## other than it's just the right magnitude to keep the mass from moving. You need another way of determining the tension in the string.
 
  • #11
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So ##F_s \ne \mu_s F_n## in that case since ##F_s = 0## but ##\mu_s F_n > 0##. The same is true in the original problem. You can't really say much about ##F_s## other than it's just the right magnitude to keep the mass from moving. You need another way of determining the tension in the string.[/QUOTE
I can't think of anything
 
  • #12
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Try thinking about the other end of the rope!
It's equal to FT acting on the hanging mass, but why can't we calculate it using the mass on the table(m1)?
 
  • #13
CWatters
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If the mass m1 isn't moving it might as well be nailed or glued in position. It makes no difference to the tension in the rope.

If the block isn't accelerating then the net force on m1 is zero. That means the actual friction force equals the applied tension force. It can't be the other way around - friction can't create tension in the rope.
 
  • #14
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If the mass m1 isn't moving it might as well be nailed or glued in position. It makes no difference to the tension in the rope.

If the block isn't accelerating then the net force on m1 is zero. That means the actual friction force equals the applied tension force. It can't be the other way around - friction can't create tension in the rope.
How do you calculate the friction force?
 
  • #15
CWatters
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Think about what vela said in #5. If the friction force acting on the block was always umg then what happens if there is no rope?
 
  • #16
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Think about what vela said in #5. If the friction force acting on the block was always umg then what happens if there is no rope?
How would there even be a friction force if there;s no rope? The only reason there's static friction is because there is an attempted motion because of the force of tension from the rope
 
  • #17
vela
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Right. The frictional force appears in response to the applied force. You can't determine the frictional force to figure out the applied force because you have to know what the applied force is to figure out what the frictional force is. That's why you have to look elsewhere to figure out the tension in the string.
 
  • #18
CWatters
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How would there even be a friction force if there;s no rope? The only reason there's static friction is because there is an attempted motion because of the force of tension from the rope
Exactly. The tension is 12*9.81 or about 120N so the actual friction force has the same magnitude. You can't calculate the actual friction force without knowing the tension because it depends on the tension.
 
  • #19
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Exactly. The tension is 12*9.81 or about 120N so the actual friction force has the same magnitude. You can't calculate the actual friction force without knowing the tension because it depends on the tension.
But if you know Fnet and you know Friction=μFn and the only 2 forces in the x component are friction and tension, then they should be equal and you should be able to do algebra to solve for tension
 
  • #20
CWatters
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But if you know Fnet and you know Friction=μFn
No, you do NOT know that the friction force is μFn.

μFn is the equation for the maximum possible friction force. You do not know that enough tension has been applied to reach the maximum friction force. Tension and friction could be less that value and it is in this problem.
 
  • #21
CWatters
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Consider what happens if m2 is variable and starts off very small and increases...

If m2=0 then the tension and friction force is also zero.
If m2 is increased to 1kg then the magnitude of the tension and friction force increases to 1*10 = 10N. The net force on m1 = 0
If m2 is increased to 2kg then the magnitude of the tension and friction force increases to 2*10 = 20N. The net force on m1 = 0
If m2 is increased to 3kg then the magnitude of the tension and friction force increases to 3*10 = 30N. The net force on m1 = 0
...
If m2 is increased to 12kg then the magnitude of the tension and friction force increases to 12*10 = 120N. The net force on m1 = 0

This continues until the tension and friction force exceeds umg = 0.45*45*10 = 202N. If m2 is increased to slightly more than 20.2kg the tension force exceeds the friction force which is limited to 202N. At this point the net force is no longer zero and the whole lot accelerates away to the right.
 
  • #22
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No, you do NOT know that the friction force is μFn.

μFn is the equation for the maximum possible friction force. You do not know that enough tension has been applied to reach the maximum friction force. Tension and friction could be less that value and it is in this problem.
OH, ok. thanks!
 
  • #23
haruspex
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this system is not in motion
Are you told that? If not, you should consider both possibilities. Since the kinetic friction is less than the static, if it is initially moving then it might continue to do so, and the tension will be less.
 
  • #24
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Are you told that? If not, you should consider both possibilities. Since the kinetic friction is less than the static, if it is initially moving then it might continue to do so, and the tension will be less.
The first part of the question, which I didn't post because I already figured it out, said that the system was in static equilibrium
 
  • #25
haruspex
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The first part of the question, which I didn't post because I already figured it out, said that the system was in static equilibrium
Ok. So do you have the answer now? If so, please Mark Solved.
 

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