Friction and Incline Planes..

[dray5150]

I'm really having trouble getting the right answer to #212 in EK 1001 Physics.

A 2kg block rest on an inclined plane with an angle of 30 degrees. A steadily increasing force is applied to the block in a direction down the inclined plane until the block begins to move. If the block begins to move when the force reaches 7.3 N, what is the approximate coefficient of static friction b/t the block and the plane?

I get that you set:
mgsin30 + 7.3N = (mgcos30)(coeffient of static friction)


I'm just not getting the EK answer of 1

Can somebody tell me what Im doing wrong?

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[MBHockey]

You have the equation right (which is the most important part) -- the coefficient of static friction is 1.007 if you solve your own equation.

What number are you coming up with? It's probably just a silly arithmetic error.

 

[dray5150]

You have the equation right (which is the most important part) -- the coefficient of static friction is 1.007 if you solve your own equation.

What number are you coming up with? It's probably just a silly arithmetic error.

static of coefficient of F = 7.8/.87 =9.0
Ugh.. what math error am I making?

 

[MBHockey]

Seems like you're missing some terms. The 0.87 is cos(30)...I'm not sure what the 7.8 is:

F + mgsin(30) = uN

F + mgsin(30) = u(mgcos(30))

u = (F + mgsin(30))/mgcos(30)

u = (7.3 + 9.8)/16.97

u = 17.1/16.97 = 1.007

 

[dray5150]

Seems like you're missing some terms. The 0.87 is cos(30)...I'm not sure what the 7.8 is:

F + mgsin(30) = uN

F + mgsin(30) = u(mgcos(30))

u = (F + mgsin(30))/mgcos(30)

u = (7.3 + 9.8)/16.97

u = 17.1/16.97 = 1.007

Thanks man you are right. I was trying to cancel out the mass and g.
Appreciate the help!!

 

[Airplay355]

g is for the acceleration due to gravity, not grams