A 5kg block rests on a 30° incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding a)up the incline, b)down the incline ?

1 Answer
Sep 25, 2017

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Let #F# be horizontal force applied on the block as shown above.This force will have two components

(1) #Fcos30^@# that will act upward parallel to inclined plane.

(2) #Fsin30^@# that will act on inclined plane perpendicularly. So it will be added with the component of the weight of the block #5gcos30^@# to enhance the magnitude of Normal reaction #N#
So the frictional force
#f_"fric"=muN=0.2(Fsin30^@+5gcos30^@)#,where acceleration due to gravity #(g) =9.8m"/"s^2#

Again the component of the weight of the block of mass 5kg acting downward parallel to the inclined plane is #5gsin30^@#

(a) Considering that the equilibrium of forces when the body is on the verge of sliding up the incline we can write

#Fcos30^@=5gsin30^@+0.2(Fsin30^@+5gcos30^@)#

#=>F=(5g(sin30^@+0.2cos30^@))/(cos30^@-0.2sin30^@)~~43N#

(b) Considering that the equilibrium of forces when the body is on the verge of sliding down the incline, we can write

#Fcos30^@=5gsin30^@-0.2(Fsin30^@+5gcos30^@)#

#=>F=(5g(sin30^@-0.2cos30^@))/(cos30^@+0.2sin30^@)~~16.6N#