Friction on a black on an inclined plane

sara_87
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Homework Statement



A block of mass 2kg is on an inclined plane and held steady. The coefficient of static friction u between the block and the plane is 0.5. Determine whether or not the block moves when released if the angle the plane makes with the horizontal is
(a) 20 degrees


The Attempt at a Solution



to find n i did this:

2*10*cos20 = 0.94

so friction is 0.47

then i did this:

2*10*sin20 - 0.47 = 6.37

so acceleration is 6.37/2 = 3.19m/s^2

so it looks like the block does move when tilted by 20 degrees...

but i feel like i got quite a big number for the acceleration, am i doing something wrong? am i supposed to approach this question differently?
 
Last edited:
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If the downward force is F and the frictional force is F_R

and [itex]F=mgsin\theta[/itex] and [itex]F_R=\mu mgcos\theta[/itex]

consider what happens if [itex]F-F_R<0[/itex] and what happens if [itex]F-F_R>0[/itex]

one says that there will not me motion and the other says that there will be

sara_87 said:
2*10*cos20 = 0.94

so friction is 0.47

then i did this:

2*10*sin20 - 0.47 = 6.37
Are you sure that is correct? I mean the frictional force part
 
Last edited:
sorry made a stupid mistake:

2*10*cos20 = 0.94

and i found a more reasonable way of doing it

the answer is supposed to be negative so the the box doesn't move
 

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