A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.333 and 0.156.
What is the minimum value of F that will prevent the block from slipping down the plane?
[tex]\sum[/tex]Fy= Sum of all Forces = Newton's Second Law
The Attempt at a Solution
So in the beginning I draw the Angles and make a Free Body Diagram. I sum up all the forces which is
[tex]\sum[/tex]Fy = [tex]\eta[/tex] + Fsin(10) - 70cos(25) = may
may is 0 since block is not moving away from the plane
what I get then is [tex]\eta[/tex]= .173648F - 63.4415
After this point I know I have to find Fsmax which equals [tex]\mu[/tex]s[tex]\eta[/tex]. After this part I get lost but I can't find [tex]\eta[/tex][tex]\mu[/tex]s
From what I remember [tex]\mu[/tex][tex]\eta[/tex]s = mg*sin[tex]\theta[/tex]. Hope somebody can help!