Answer
The minimum force needed to hold the picture static against the wall is $16.33$ $N$
Work Step by Step
Let $ F$ be the applied force on the picture and $m$ be the mass of the picture.
Now the forces acting on $m$ are
(a) weight ($mg$) of the picture in downward direction,
(b) friction ($f$) in upward direction,
(c) normal force ($N$) in outward direction from the wall,
(d) external force ($ F$) towards the wall
As the picture is in static condition, total resultant forces acting on the picture would be zero.
Thus, $f=mg$ and $N=F$
Again, we know that $f=\mu N$, where $\mu$ is the coefficient of static friction
between the picture and the wall.
Thus,
$mg=\mu F$
or, $F=\frac{mg}{\mu}$
Putting $m=1.10$ $kg$, $g=9.8$ $m/s^{2}$, and $\mu =0.660$, we get,
$f=16.33$ $N$
Therefore the minimum force needed to hold the picture static against the wall is $16.33$ $N$
