Static equilibrium: A cubic block is on an incline....

[Timcago]

A cubic block is sitting at rest on an incline, as drawn. The block weighs 10 lb. The normal force acting on the black is 8.1 lb. What is the magnitude of the frictional force, F, holding the block in place? Also, write the frictional force as a vector, using the unit components I and J. (this is an application of static equilibrium)

 

[skeeter]

The system is in equilibrium, which means F + N + W = 0.

Translate the force of static friction, F, downward so that the head of F touches the tail of N and you end up with a right triangle.

now use Pythagoras to determine F ...

F² = W² - N²

... then you can determine F in component notation by figuring out a few angles.

 

[Timcago]

When i shift F down i get a 90 degree angle with w in between.

So i should shift both F and W at the same time so that you have a triangle with the angle between N and f being 90 degrees.

Then using the F^2 = W^2 - N^2, you get 5.86 for F. Is that the magnitude?

Is this what i should do next?

Find an angle using the law of cosines and then find the 3rd angle by subtracting the other 2 angles by 180?

Next i should find the Horizontal and vertical components of 1 along the other?

This is where i am stuck. How do i find the component notation?

 

[Timcago]

I got that the angle between F and W is 54.1 degrees and the angle between N and W is 35.9 degrees.

Where do i go from here knowing F is 5.86

Horizontal component i do 5.86*cos(90) = 0
Vertical component 5.86*sin(90)= 5.86

F= 0I + 5.86J?

 

[skeeter]

the angle between N and W (35.9 degrees) is the same as the angle of the incline with respect to the horizontal.

using the horizontal as 0, the force of friction, F, points toward quadrant II.
so, in component form ...

F = -5.86cos(35.9)i + 5.86sin(35.9)j