a block of mass 5 kg is placed on an inclined plane at 45 to the horizontal. wha

Question

a block of mass 5 kg is placed on an inclined plane at 45 to the horizontal. what is the minimum coefficient of static friction so that that block remains at rest? I think it's 1, but I dont have access to the answer and I want to check my work.

I know there is no net force in the y direction so Fn=mg*cos(45).
There is also no net force in the x direction, so f (frictional force)=mg*sin(45)
The frictional force= the coefficient of static friction times the normal force.
To solve for coeff. static friction:
(mg*sin(45))/mg*cos(45)
mg cancels out, and sin(45) =.5 and cos(45) also =.5
.5/.5=1

is this the right process?

Explanation / Answer

The process is right. The minimum friction coefficient can also be calculated as =Tan-1()=Tan-145=1

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