Forces: incline+static friction

Question

The figure below shows a 20-kg block on a fixed incline that makes a fixed angle of 60

with the ground. A constant wind force F is blowing up, and along the incline, as indicated

in the figure.

The block rests on the incline and the wind

force (F) blows up along the incline as shown

If the coefficient of static friction between the block and the incline is us =0. 6,

a) determine the range of values for F so that the block remains in static

equilibrium.

b) Determine the value for F so that the block remains in static equilibrium even if

there is no friction between the block and the incline

Answer 1

There is no figure given, but the problem can be worked from the description. Let’s define the forces involved.

Fd = force on block parallel to the incline and downward = M x g x sin60 = 170 N

Fr = static friction force = M x g x cos60  x 0.6 = 59 N (M x g cos60 = normal force)

Fw = force on block due to the wind.

For the block to be in static equilibrium, Fd = Fw + Fr since Fw and Fr oppose the force pushing the block downward.

So, Fw = Fd – Fr or Fw = 111 N.

If we neglect the frictional force, then Fw = Fd