Static friction preventing block from sliding

[malta]

Homework Statement

If the coefficient os static friction between the incline and block is .3, then what is the minimum force,F, needed to prevent the block from sliding?
m=20kg theta= 50 degrees

Homework Equations

Normal force= mgcosx
Fs= uk mgcosx

The Attempt at a Solution

I know how to solve this this for the perpendicular force (up the ramp) but I can't seem to figure out how to do it for thr parallel force.

 

[AEM]

Homework Statement

If the coefficient os static friction between the incline and block is .3, then what is the minimum force,F, needed to prevent the block from sliding?
m=20kg theta= 50 degrees

Homework Equations

Normal force= mgcosx
Fs= uk mgcosx

The Attempt at a Solution

I know how to solve this this for the perpendicular force (up the ramp) but I can't seem to figure out how to do it for thr parallel force.

I assume that you drew yourself a diagram with a coordinate system that has its x-axis parallel to the inclined plane and its y-axis perpendicular to the plane. Then, what is the force due to gravity acting to move the block down the plane? What is the frictional force acting to oppose the motion down the plane? What is the condition on the sum of all forces along the plane so that the block does not move?

 

[malta]

Fore due to gravity is mgsinx and the frictional force = (coefficient of fiction)(mgCosx)
And the condition on all forces along the plane so that the block doesn't move has to equal to zero

But I can't seem to get what to set equal to what

 

[malta]

anyone? I am doing a test review for a test tomorrow so this would greatly help me

 

[LowlyPion]

Fore due to gravity is mgsinx and the frictional force = (coefficient of fiction)(mgCosx)
And the condition on all forces along the plane so that the block doesn't move has to equal to zero

But I can't seem to get what to set equal to what

You have the right parts.

Draw a diagram and add the forces that need to make it be static.

 

[malta]

ok so
F + mgcosx= N
(friction)N= mgSinx

So F = (mgSinx)/(friction) - mgCosx

There we go, thanks again

 

[LowlyPion]

ok so
F + mgcosx= N
(friction)N= mgSinx

So F = (mgSinx)/(friction) - mgCosx

There we go, thanks again

Not exactly.

F + μmgcosθ = mgsinθ

You only need to supply force in excess of what friction alone will not.