# Calculating the coefficient of friction Given 2 forces and and angle

## Homework Statement:

A force, FA, of 8.70 kN is the maximum force that can be applied to a 2.48 kN box before it starts to move. If the force, FA, is angled at 10.0° above the horizontal

## Relevant Equations:

was not given the equation to do this kind of problem or so i believe the one i was given is Fc+fn=Fg
i tryed putting all the number into that equation but it does not work, i have no clue if i need to include the angle to solve this or not and if i do i have no clue how i would

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You need to show your work before we can really help you. However since you are at that stage perhaps this video from Khan will help get you started:

My work- so far i have tryed to use Fk/Fn=coefficient of static imputing my numbers as 8700/2480=3.50 and 2480/8700=0.29 but neither of these give the correct answer, i dont understand where the angle should be put into my calculation

jedishrfu
Mentor
the force pushes your block from an angle so you need to compute the horizontal component that is pushing it and the vertical that is pushing down on it.

Watch the video and see if that helps.

• HTML
SammyS
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My work- so far i have tried to use Fk/Fn=coefficient of static imputing my numbers as 8700/2480=3.50 and 2480/8700=0.29 but neither of these give the correct answer, i don't understand where the angle should be put into my calculation
You likely have seen an expression for static friction which is something like ##F_\text{Static} \le \mu _\text{Static} \cdot F_\text{Normal} ## .
You were given the maximum force that can be applied before movement occurs, so equality holds.

It appears that you did not give a complete statement of the problem. (I assume that the box is on a horizontal surface.)

How is the force of friction related to the applied force, FA ? (The 10° angle is important here.)

How is the normal force to be determined?

• jedishrfu
You likely have seen an expression for static friction which is something like ##F_\text{Static} \le \mu _\text{Static} \cdot F_\text{Normal} ## .
You were given the maximum force that can be applied before movement occurs, so equality holds.

It appears that you did not give a complete statement of the problem. (I assume that the box is on a horizontal surface.)

How is the force of friction related to the applied force, FA ? (The 10° angle is important here.)

How is the normal force to be determined?
yes the box is on a horizontal slope at 10 degrees and i dont know how the normal force is determined
fn=fg but i was not given a mass in the question so i dont get how to determine it without a mass

jedishrfu
Mentor
@HTML I reiterate you need to think harder on this. We can’t do your homework for you. We can’t walk you through the solution. If you show us your work then we can say whether it’s right or not and if not why with a hint to push you in the right direction.

Did you look at the Khan Academy video yet? It should have given you the basics here.

@HTML I reiterate you need to think harder on this. We can’t do your homework for you. We can’t walk you through the solution. If you show us your work then we can say whether it’s right or not and if not why with a hint to push you in the right direction.

Did you look at the Khan Academy video yet? It should have given you the basics here.
i did and it mentioned absolutely nothing about angles i under stand the basics i just do not under stand why and where the angle should come into play i have tried another formula- fsMax/FnCos theata but its still not working i used this formula because i believed that my fs max would be the 8.73 and the fn would be the 2.48 since it is acting in the downward direction to keep the box on the surface.

SammyS
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@HTML ,
If the weight of the box is 2.48 kN, then what is its mass?

haruspex
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fn=fg
No.
Consider all the forces acting that have a vertical component. Write the force balance equation for the vertical direction. Then do the same for the horizontal direction.

@HTML ,
If the weight of the box is 2.48 kN, then what is its mass?
252.8 kg? i took the 2.48kN and put it into 2480 N then divided it by 9.81 and got the 252.8

No.
Consider all the forces acting that have a vertical component. Write the force balance equation for the vertical direction. Then do the same for the horizontal direction.
thats where im confused because of the angle in the x direction there is fn and fg and fa but at a 10 degree angle and in the y direction there is fs and fa at a angle so where does the 10 degrees go do i need to split the fa into its x and y components

haruspex
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do i need to split the fa into its x and y components
Yes.
in the x direction there is fn and fg
in the y direction there is fs and fa at a angle
It is more usual to use y for vertical and x for horizontal.

Yes.

It is more usual to use y for vertical and x for horizontal.
shouldent it just be sin theata/ cos theata= tan theta, tan theata=coeficcent of static

haruspex
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shouldent it just be sin theata/ cos theata= tan theta, tan theata=coeficcent of static
Shouldn't what be tan theta?
The coefficient of static friction turns out to be tan theta in a certain common problem, but it is not in general true.

What are the horizontal and vertical components of the applied force?

horizontal components would be Fs and vertical would be Fn and Fg? correct?

haruspex
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horizontal components would be Fs and vertical would be Fn and Fg? correct?
What are the horizontal and vertical components of the applied force FA?

SammyS
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252.8 kg? i took the 2.48kN and put it into 2480 N then divided it by 9.81 and got the 252.8
That does give the mass (as long as the unit of measure you omitted is kg), although, you don't need to know the mass to do this problem.

One thing which would be very helpful here would be a good sketch of the problem. Better yet would be to use an F.B.D. (free body diagram).

What are the components of the applied force, FA ?

SammyS
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that's where I'm confused because of the angle in the x direction there is fn and fg and fa but at a 10 degree angle and in the y direction there is fs and fa at a angle so where does the 10 degrees go do i need to split the fa into its x and y components
I see that @haruspex replied to this previously, but I will add my 2 cents worth.

As I stated previously, a Free Body Diagram, commonly abbreviated as "fbd",will help you solve the problem addressed in this thread. In general, FBDs are extremely useful and often necessary to solve statics and dynamics problems in physics and engineering. Here is a link to the Wikipedia page on FBDs.

Here is an FBD for your problem. The vectors are not drawn to scale and the angle, θ, is not 10° .

FS is the force of static friction.
FN is the Normal force (exerted on the box by the supporting horizontal surface).
W is the weight of the box, what you referred to as fg in post # 16.
FA is the applied force. The inset in the above figure shows FA broken down into its horizontal (x) and vertical (y) components.