Static Friction Force

[soby10]

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I have question from TPR Physics.

1. A box of mass m is sitting on an incline of 45 degrees and it requires an applied force F up the incline to get the box to begin to move. What is the maximum coefficient of static friction?

I know that the parallel component of the gravitational force is mgsin45 and the perpendicular component is mgcos45. The maximum static frictional force would be (normal force mgcos45)*the static coefficient. The question asked about when the box is just about to move so the upward and downward forces are equal and static friction is at a max. But the explanation states F=static coefficient*mgcos45+static coefficient*sin45 and from there you can solve for the coefficient. Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

answer is {2F^(1/2)/(mg)}-1

 

[Rabolisk]

Not quite the answer I got.

 

[ModusProbandi]

Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

Precisely.

 

[cartman1980]

I know that the parallel component of the gravitational force is mgsin45 and the perpendicular component is mgcos45.

Perhaps my conceptual understanding is skewed up. If someone would be kind enough to point it out, it'd be much appreciated

If i draw the picture in my head,
- mgsin45 is upwards along the slope
- frictional force (mew)mgcos45 is opposing the motion of the block

at maximum static friction, both these are equal

mg sin45 = (mew) mg cos45
(mew) = tan 45

 

[Rabolisk]

Perhaps my conceptual understanding is skewed up. If someone would be kind enough to point it out, it'd be much appreciated

If i draw the picture in my head,
- mgsin45 is upwards along the slope
- frictional force (mew)mgcos45 is opposing the motion of the block

at maximum static friction, both these are equal

mg sin45 = (mew) mg cos45
(mew) = tan 45

Why?

 

[cartman1980]

yes, i see the obvious flaw... had the direction slightly mixed up when drawing in my head.. paper drawing seems to indicate

- F - mgsin45 along the slope
- (mew) mgcos45 opposing the motion

F - mgsing45 = (mew)mgcos45 is the equation then?

 

[justhanging]

why do you multiply the static friction by sin45 if that is the gravitational force pulling the box down, shouldn't it be mgsing45?

 

[Rabolisk]

why do you multiply the static friction by sin45 if that is the gravitational force pulling the box down, shouldn't it be mgsing45?

I don't think anyone multiplied the static friction by sin45, and I'm not exactly sure I understand what you are asking.

 

[justhanging]

I have question from TPR Physics.

1. A box of mass m is sitting on an incline of 45 degrees and it requires an applied force F up the incline to get the box to begin to move. What is the maximum coefficient of static friction?

I know that the parallel component of the gravitational force is mgsin45 and the perpendicular component is mgcos45. The maximum static frictional force would be (normal force mgcos45)*the static coefficient. The question asked about when the box is just about to move so the upward and downward forces are equal and static friction is at a max. But the explanation states F=static coefficient*mgcos45+static coefficient*sin45 and from there you can solve for the coefficient. Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

answer is {2F^(1/2)/(mg)}-1

Quote:
Originally Posted by soby10
Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

Precisely.




Don't know why this person put precisely, unless am missing something. Doesn't make sense should be mg not static coefficient.

 

[Rabolisk]

I don't know either. It should be mg.

 

[lolarogers]

The static friction force must be overcome by an applied force before an object can move. The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force:

.

 

[soby10]

I thought it would be mg also but according to TPR Physics and Math Review 2010 it's not. I'm a bit confused about the answer. Has anyone worked out this prob? The explanation states F=static coefficient*mgcos45+static coefficient*sin45. I thought it shouldn't be static coefficient but mg. Is this a typo?