Coefficient of friction along an inclined plane

AI Thread Summary
A block on an inclined plane requires a horizontal force of 2N to prevent it from sliding down, and the incline is at an angle of 16.3 degrees. The coefficient of friction was initially miscalculated, but the correct formula led to a value of approximately 0.083. The user struggled with calculating acceleration after removing the force, mistakenly combining units and failing to account for mass in their calculations. Emphasis was placed on the importance of using consistent units and drawing free body diagrams for clarity. Despite resolving the coefficient of friction, confusion remained regarding the acceleration calculation.
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.
 
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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
 
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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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