A block of mass m=2kg is kept in position on the wall by applying an oblique force 'F' at 60° to the vertical as shown . calculate the minimum value of 'F'(the coafficient of friction between the wall and the block = 0.9) (Take 'g'=10ms^-2)?

1 Answer
P dilip_k
Jul 7, 2018

enter image source here

Let the force #vecF# be applied on the block obliquely at an angle of #theta=60^@# with the vertical as shown in the figure above.

The resolved part of the applied force #vecF# perpendicular to the wall will be #vecFsintheta# and the resolved part parallel to the wall in upward direction is #vecFcostheta#

So static fictional force in upward direction is

#F_suarr="coefficient of friction"(mu_s)xxFsintheta#

#=0.9vecFsintheta#

So considering the equilibrium of forces we have

#vecFcostheta+0.9*vecFsintheta=mg#

#=>vecFcos60+0.9*vecFsin60=2*10#

#=>vecF=20/(1/2+(0.9sqrt3)/2)~~15.6N#

If the applied force #vecF# acts obliquely downward making an angle #theta=60^@# with the vertical then the component of #vecF# in vertical direction will be downward and equilibrium of force will give the following relation.

#0.9*vecFsintheta=mg+ vecFcostheta#

#=>-vecFcos60+0.9*vecFsin60=2*10#

#=>vecF=20/(-1/2+(0.9sqrt3)/2)~~71.6N#

Related questions
Impact of this question
2855 views around the world
You can reuse this answer
Creative Commons License