What is the minimum tension in the string?

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The discussion centers on calculating the minimum tension in a string holding a 15kg block stationary on a 22-degree incline with a coefficient of friction of 0.12. Participants emphasize the importance of understanding static friction, noting that its force can vary, affecting the tension required to keep the block at rest. The minimum tension occurs when static friction is at its maximum, allowing for the calculation of forces acting on the block. After resolving some calculation errors related to using radians instead of degrees, the final calculation yields a minimum tension force of approximately 39 N. The conversation highlights the significance of accurately applying physics principles and ensuring correct unit usage in calculations.
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Homework Statement


A 15kg block sitting on a 22(degree) incline is held stationary by a string as shown. The coefficient of friction between the block and the surface of the incline is 0.12.

What is the minimum tension in the string?

Homework Equations


F=mg

The Attempt at a Solution


Drew a free-body diagram.http://postimg.org/image/dxczxynad/
 
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Not certain, but if I calculate the Fgparallel portion, will I be able to calculate my Tension Force?
 
Ryan Lau said:
Not certain, but if I calculate the Fgparallel portion, will I be able to calculate my Tension Force?
Yes that will be necessary, but then what? Consider a free body diagram, what are all the forces acting on the block? Two of them are variable.

Edit:
I didn't see you already drew one. Which of the forces are variable? (Hint: their sum is not variable)
 
Well, I notice that both Tension Force and Force Friction are along the same path, therefore I believe those are the two variables?
 
It is stationary thus the net sum of forces acting it is zero.
 
Ryan Lau said:
Well, I notice that both Tension Force and Force Friction are along the same path, therefore I believe those are the two variables?
Well, yes. But not for your reasoning. The reason is that the force of static friction is variable. μsFn only represents the maximum force that static friction can apply, but it is possible for static friction to apply less or even no force at all.

You're asked to find the minimum tension in the rope. When do you think the force of tension will be minimum?
When the static friction is zero? When the static friction is maximum? Or somewhere in between?
 
Nathanael said:
Well, yes. But not for your reasoning. The reason is that the force of static friction is variable. μsFn only represents the maximum force that static friction can apply, but it is possible for static friction to apply less or even no force at all.

You're asked to find the minimum tension in the rope. When do you think the force of tension will be minimum?
When the static friction is zero? When the static friction is maximum? Or somewhere in between?
I noticed you mentioned static friction.

Since the object is at rest, I believe that the static friction is somewhere in between.
 
Ryan Lau said:
Since the object is at rest, I believe that the static friction is somewhere in between.
It can be somewhere in between. Or it can be maximum. Or it can be zero. All of these are valid possibilities. But for each of these options, the force of tension will be different.
When will the force of tension be minimum? Are you saying tension is minimum when the force of friction is somewhere in between it's minimum (zero) and it's maximum?
 
  • #10
Nathanael said:
It can be somewhere in between. Or it can be maximum. Or it can be zero. All of these are valid possibilities. But for each of these options, the force of tension will be different.
When will the force of tension be minimum? Are you saying tension is minimum when the force of friction is somewhere in between it's minimum (zero) and it's maximum?

Tension force is minimum when force of friction is at it's maximum.
 
  • #11
Ryan Lau said:
Tension force is minimum when force of friction is at it's maximum.
Good. So now find the minimum force of tension. (That is, find the tension needed to keep the object stationary when static friction is applying it's maximum force.)
 
  • #12
Nathanael said:
Good. So now find the minimum force of tension. (That is, find the tension needed to keep the object stationary when static friction is applying it's maximum force.)

Okay, I think I may be getting what you mean, however I end up with a diagram like this. Is there anything in particular that I am doing incorrect?

http://postimg.org/image/nujh0ea99/

http://postimg.org/image/y9ir71kwn/
 
  • #13
How do you get Fgravity parallel=64.7 N and Fgravity perp.=147 N?
 
  • #15
Ok you have the right idea, but 64.7 is not the right number.

Can you finish solving for the minimum tension now?
 
  • #16
Nathanael said:
Ok you have the right idea, but 64.7 is not the right number.

Can you finish solving for the minimum tension now?

147cos68 = 64.70102431 N
^ This is my adjacent, a.k.a. my Gravitational Force parallel. Should the opposite of it not equal to both the Tension Force and Frictional Force combined? I thought that is how equilibrium works.

I am having difficulty finding the correct formula to solve. My assumptions are that (0.12 Mew * Normal Force) is my maximum Frictional Force? The issue is also found when I try to find the adjacent for the 22 degree triangle.
 
  • #17
Ryan Lau said:
147cos68 = 64.70102431 N
This is not true... You need to make sure your calculator is in "degrees mode" and not radians.

Ryan Lau said:
Should the opposite of it not equal to both the Tension Force and Frictional Force combined? I thought that is how equilibrium works.
Yes this is correct.

Ryan Lau said:
My assumptions are that (0.12 Mew * Normal Force) is my maximum Frictional Force?
Yes this is also correct.

Ryan Lau said:
The issue is also found when I try to find the adjacent for the 22 degree triangle.
I'm not sure I understand what you're confused about. Are you having trouble finding the normal force?
 
  • #18
Nathanael said:
This is not true... You need to make sure your calculator is in "degrees mode" and not radians.Yes this is correct.Yes this is also correct.I'm not sure I understand what you're confused about. Are you having trouble finding the normal force?
Oh my heavens...

Mr. Nathanael, I am terribly sorry... This entire time I was using radians!
Brief overview.

Force of Gravity Parallel = 147cos68 = 55.0672 N
Force of Gravity Perpendicular = 147cos22 = 136.2960 N
Force of Friction = (Mew * Normal Force) = (0.12)(136.2960) = 16.35552

MINIMUM TENSION FORCE = Force of Gravity Parallel - Force of Friction = (55.0672) - (16.35552) = 38.71168 = 39 N
 
  • #19
Ryan Lau said:
Oh my heavens...

Mr. Nathanael, I am terribly sorry... This entire time I was using radians!
Brief overview.

Force of Gravity Parallel = 147cos68 = 55.0672 N
Force of Gravity Perpendicular = 147cos22 = 136.2960 N
Force of Friction = (Mew * Normal Force) = (0.12)(136.2960) = 16.35552

MINIMUM TENSION FORCE = Force of Gravity Parallel - Force of Friction = (55.0672) - (16.35552) = 38.71168 = 39 N
Good job, it is correct.
 
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