Need h/w help asap

[ahmed]

A force p, parallel to a surface inclined 15 deg. above the horizontal, acts on a 45-N block, as shown below in the attached pic.
The coefficients of friction for the block and surface are static u_s=0.5 and u_k=0.34. If the block is initially at rest,
determine the magnitude and direction of the frictional force acting on the block for magnitudes of p of
a) 5N
b)8N
c)15N

[Mao]

the block weighs 45 newtons. At an angle of inclination of 15 degrees, it has N parallel to the surface, and N perpendicular to the surface. To overcome static friction, the sum of forces in the direction of motion must be greater than maximum static friction. Since P is acting downhill, static friction must act uphill.

Since P is acting in the horizontal (i) direction, it has a component parallel and perpendicular to the plane as well.

Resolving forces perpendicular to the plane, P and the normal reaction force N pulls the block away from surface, gravity pulls down:



Resolving forces parallel to the plane

If the total down-slope forces are greater than static friction:
Then , where resistive forces

Otherwise, (total resistive force is equal to total downslope force, no net motion)

sub in the values of P to see which of the above two scenario it fits.