The free-body diagram of the given situation is shown in the following figure. The balancing force in x and y directions is given by N+Ncos60∘=mg⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1) That is, Nsin60∘=f⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)
From Eq. (1), we get N=
16
1+(1∕2)
=
32
3
From Eq. (2), we get the frictional force at the bottom of the stick as follows: (
32
3
)(
√3
2
)=f⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(3) ⇒f=
16√3
3
N From triangle ABC, we get
h
x
=sin60∘ ⇒x=
h(2)
√3
⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(4) Now, the balancing torque about the bottom of the stick is Nx−(