Isn't the additional force of 1 N pretty arbitrary? Wouldn't the budging force be just over 49, so the coefficient of friction would be 49/49sqrt3=1/sqrt3?

This was completely confusing me as well: how did Sal come up with 1N? All he's saying is that with the hypothetical materials of the block and plane (which are completely unknown really) he began applying force and at 0.5N it didn't move, at 0.99N it didn't move, but finally at exactly 1N the block began to move. So in this case the coefficient of static friction was 49N at rest + the 1N extra force to finally make it move = 50N (or budging force) and divide that by the normal force to get coef of static friction of 0.59 for this scenario's materials. In other words, he could've just as easily said it took 10N to finally move the object. He tried 5N, then 7N, 9.99N and finally at 10N the object moved. in this case the budging force would've been 59N (the 49N of parallel force + the extra 10N of force to finally get the block to move) and the coef would've been 59N / 49*sqrt3N = mu(static) or 0.695.

what is coefficient? what is it's usage in this question? thanks

This is the "coefficient of static friction". It is a dimensionless measure of how much frictional force exists between an object and a surface. It indicates the force resisting the object to get started moving. The coefficient of static friction is typically larger than the coefficient of kinetic friction, which indicates the friction force that exists while the object is moving.

Wait, isn't the coefficient of static friction the tan of theta? That is what my textbook says. So shouldn't the coefficient of static friction be tan(theta) = tan(30) = 0.577 ?

No. What your textbook probably says, that an angle theta for which a block will begin to move spontaneously under its own weight on an inclined plane, must be related to the coefficient of friction between block and the incline, mu, as mu=tan(theta). That is called the angle of repose. mu is a property of the nature of surfaces in contact only.

Why did he add additional 1N? Can't he add .5N (or any other number) because that is still greater than 40N?

Yes, he could have. Adding 1N just made the math easier to handle than a decimal or something large 1,546,785.

I'm still very confused on the "budging force" example. Normally when asked to find the coefficient for static friction you are only told so much information. how do we come up with the "budging force" given limited information. 1N seemed pretty arbitrary. Please help!

Is static friction a constant? I am confused as to why it was 1N when it could have been 0.1. Wouldn't that change the static friction? Is the static friction different for different materials, and how would that be calculated or would it have to be given?

what are the units of the coefficient of static friction?

It is unitless, otherwise friction would have units different from the normal force, so it would not be a force.

Main content

Physics library

Course: Physics library > Unit 3

Lesson 5: Inclined planes and friction

Correction to force of friction keeping the block stationary

Name: Correction to force of friction keeping the block stationary
Uploaded: 2011-06-20T23:26:40Z
Description: Correction to Force of Friction Keeping the Block Stationary

AP.PHYS:

INT‑3.B.2 (EK)

Google Classroom

Correction to Force of Friction Keeping the Block Stationary. Created by Sal Khan.

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Mrbobthebuilder6
Posted 10 years ago. Direct link to Mrbobthebuilder6's post “Isn't the additional forc...”
Isn't the additional force of 1 N pretty arbitrary? Wouldn't the budging force be just over 49, so the coefficient of friction would be 49/49sqrt3=1/sqrt3?
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(8 votes)
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- JJ Arnold
  Posted 9 years ago. Direct link to JJ Arnold's post “This was completely confu...”
  This was completely confusing me as well: how did Sal come up with 1N?
  
  All he's saying is that with the hypothetical materials of the block and plane (which are completely unknown really) he began applying force and at 0.5N it didn't move, at 0.99N it didn't move, but finally at exactly 1N the block began to move. So in this case the coefficient of static friction was 49N at rest + the 1N extra force to finally make it move = 50N (or budging force) and divide that by the normal force to get coef of static friction of 0.59 for this scenario's materials.
  
  In other words, he could've just as easily said it took 10N to finally move the object. He tried 5N, then 7N, 9.99N and finally at 10N the object moved. in this case the budging force would've been 59N (the 49N of parallel force + the extra 10N of force to finally get the block to move) and the coef would've been 59N / 49*sqrt3N = mu(static) or 0.695.
  Comment on JJ Arnold's post “This was completely confu...”
  (15 votes)
Morwarid Najafizada
Posted 9 years ago. Direct link to Morwarid Najafizada's post “what is coefficient? what...”
what is coefficient? what is it's usage in this question? thanks
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(2 votes)
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- George Arrington
  Posted 9 years ago. Direct link to George Arrington's post “This is the "coefficient ...”
  This is the "coefficient of static friction". It is a dimensionless measure of how much frictional force exists between an object and a surface. It indicates the force resisting the object to get started moving. The coefficient of static friction is typically larger than the coefficient of kinetic friction, which indicates the friction force that exists while the object is moving.
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  (7 votes)
Xihai Luo
Posted 10 years ago. Direct link to Xihai Luo's post “Wait, isn't the coefficie...”
Wait, isn't the coefficient of static friction the tan of theta? That is what my textbook says. So shouldn't the coefficient of static friction be tan(theta) = tan(30) = 0.577 ?
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(2 votes)
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- Satwik Pasani
  Posted 10 years ago. Direct link to Satwik Pasani's post “No. What your textbook p...”
  No. What your textbook probably says, that an angle theta for which a block will begin to move spontaneously under its own weight on an inclined plane, must be related to the coefficient of friction between block and the incline, mu, as mu=tan(theta). That is called the angle of repose. mu is a property of the nature of surfaces in contact only.
  Comment on Satwik Pasani's post “No. What your textbook p...”
  (6 votes)
Awsaf Alam Anindo
Posted 11 years ago. Direct link to Awsaf Alam Anindo's post “why is energy a scalar qu...”
why is energy a scalar quantity
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(4 votes)
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- azuro010
  Posted 10 years ago. Direct link to azuro010's post “though energy is a vector...”
  though energy is a vector quantity but still from experience we can say that it works in a definite direction. But it is scalar quantity because the final product of Mass*square of velocity is always a positive quantity. Hence, by mathematical calculations , energy need not be scalar.
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  (0 votes)
Julia
Posted 8 years ago. Direct link to Julia's post “Why did he add additional...”
Why did he add additional 1N? Can't he add .5N (or any other number) because that is still greater than 40N?
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(2 votes)
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- Marissa Tanner
  Posted 7 years ago. Direct link to Marissa Tanner's post “Yes, he could have. Addi...”
  Yes, he could have. Adding 1N just made the math easier to handle than a decimal or something large 1,546,785.
  Comment on Marissa Tanner's post “Yes, he could have. Addi...”
  (3 votes)
Christian LaDon
Posted 8 years ago. Direct link to Christian LaDon's post “I'm still very confused o...”
I'm still very confused on the "budging force" example. Normally when asked to find the coefficient for static friction you are only told so much information. how do we come up with the "budging force" given limited information. 1N seemed pretty arbitrary. Please help!
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(1 vote)
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- samimorin654
  Posted 7 years ago. Direct link to samimorin654's post “Is static friction a cons...”
  Is static friction a constant? I am confused as to why it was 1N when it could have been 0.1. Wouldn't that change the static friction? Is the static friction different for different materials, and how would that be calculated or would it have to be given?
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  (2 votes)
Harvey G
Posted 8 years ago. Direct link to Harvey G's post “What is the difference b...”
What is the difference between static and kinetic friction
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(2 votes)
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- N.Fajardo28
  Posted 8 years ago. Direct link to N.Fajardo28's post “Static Friction is the fr...”
  Static Friction is the friction present in a stationary object. In order to set a object in motion, the applied force must be greater than the force of static friction. Kinetic friction on the other hand is the force of friction present when a object is moving. Static friction will become kinetic friction once the object has been set in motion.
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  (1 vote)
Wizzy.Wig16
Posted 8 years ago. Direct link to Wizzy.Wig16's post “what are the units of the...”
what are the units of the coefficient of static friction?
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(1 vote)
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- roht2326
  Posted 8 years ago. Direct link to roht2326's post “It is unitless, otherwise...”
  It is unitless, otherwise friction would have units different from the normal force, so it would not be a force.
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  (3 votes)
Rafea Taji
Posted 7 years ago. Direct link to Rafea Taji's post “I think Sal used 1 N addi...”
I think Sal used 1 N additional force because he wanted to just overcome the static friction. Had it been a greater additional force, coefficient of Kinetic friction would have come into action. Correct me if I'm wrong.
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(2 votes)
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- Daniel Santos
  Posted 7 years ago. Direct link to Daniel Santos's post “Yes. It was only a hypoth...”
  Yes. It was only a hypothetical scenario so that Sal can show us how a budging force affects the 'coefficient of static friction'.
  Comment on Daniel Santos's post “Yes. It was only a hypoth...”
  (1 vote)
dawood.aijaz97
Posted 7 years ago. Direct link to dawood.aijaz97's post “how to calculate gravitat...”
how to calculate gravitational potential energy of a mass on inclined plane of height H?
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Video transcript

Just realized that I typed in the wrong numbers at the end of the last video when I was trying to put them in the calculator. We were trying to determine the coefficient of static friction, and it's 50 newtons divided by 49 square roots of 3 newtons. And in the last video, by accident, in the calculator, instead of doing 49 square roots of 3, I wrote 40 square roots of 3. So let's actually calculate it again. 50 divided by 49 times-- let me write it this way, make sure we get-- 49 times the square root of 3 gives us a coefficient of static friction of, if we round it, 0.59. So this is equal to 0.59. And so that's it. I just forgot to-- I did the calculation wrong on the calculator in the last video.