Coefficient of friction along an inclined plane

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SUMMARY

The discussion focuses on calculating the coefficient of friction and acceleration for a block on an inclined plane with a 16.3-degree angle. A horizontal force of 2N is applied to prevent the block from sliding down a rough surface. The correct coefficient of friction is determined to be approximately 0.083, while the acceleration when the force is removed is calculated using the equation F=ma. Key errors identified include incorrect unit handling and the necessity of incorporating mass in calculations.

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Ulti
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Homework Statement


A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin 7/25 (=16.3 degrees) to the horizontal. Find the coefficient of friction between the block and the plane and then acceleration with which the block will move when the force is removed.

Homework Equations


Friction = Coefficient of friction x Resistive force (parallel at a right angle to the slope on which the objects acts on)
Force = Mass x Acceleration

The Attempt at a Solution


Well what I thought was that the force that the object has should be 2+gsin16.3 which equals 4.75N. This means that the acceleration will be 2.75N when the force of 2N is taken away. However the answer is meant to be 1.97ms^-2... Did I work out the acceleration wrong in the first place?

Now attempting to work out the coefficient of friction, I simply did friction over resistive force: gsin16.3/gcos16.3 and that gives me 0.292 but the answer is supposed to be 0.0827... I do not understand what I am doing wrong.

If someone can say what I'm doing wrong and lead me into the right direction then thanks!

EDIT: Worked out coefficient of friction now but can't work out acceleration.
 
Last edited:
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check your units in all your calculations. Sometimes you're missing critical variables and units will help you.

for instance your statement:
"2 + g sin16.3 which equals 4.75N."
is mathematically off in terms of units.

You have the applied force (2N) plus 9.8 m/s^2 times a quantity that doesn't have units. why is this bad: You can't add things with different units together!

then you have this:
"acceleration will be 2.75N"
is that right in terms of units?

Note: units will help you a LOT! So will a free body diagram. It looks, however, that you're headed the right direction. :smile:
 
Ah I see, so that's where I went wrong on that part, I've drawn a free body diagram but I always thought I could just add the 2N onto it as it is and this is the first time I encountered a problem. (Obviously I made a big mistake when I said the acceleration was measured in N xD)

EDIT: Can't seem to work out the force... Not sure how to turn the 2N force into the same unit... Really confused now :S
 
Last edited:
Wow I solved the coefficient of friction...
It was (g sin16.3 - 2 cos 16.3)/(-2 sin 16.3 - g cos 16.3) = 0.083

No idea how to work out acceleration still though.
 
Ulti said:
Wow I solved the coefficient of friction...
It was (g sin16.3 - 2 cos 16.3)/(-2 sin 16.3 - g cos 16.3) = 0.083

No idea how to work out acceleration still though.

Use F=ma... Force (in N) relates to mass (in kg) times acceleration (in m/s^2).

When you have your free body diagram, you need to have the weight force parallel to the slope as m*g*sin theta... you left out the mass.

Always check units!: in this case: N = kg*m/s^2
 
Well as mass is 1kg, I can basically ignore it I think. I am correct in thinking that? Still not sure how the units apply here...
 
Sigh.. still no luck after a few hours. I don't know how many hours I've spent in total on this...
 
Bump?
 

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