'Ampleness' of a big line bundle Let $X$ be a non-singular complex variety with a big line and base point free bundle $M$ on it. My question is can we say that for any locally free sheaf $F$ on $X$, $F \otimes M^n$ is globally generated for $n \gg 0$. 
Motivation: If $M$ were an ample line bundle then all we need is that $F$ is coherent sheaf. But since we are given a much stronger condition on $F$ (which is local freeness) can we say the same thing with $M$ just being big and base point free.
I tried to use the fact that any big line bundle is tensor product of an ample line bundle and an effective line bundle. 
I am not even sure that this has to be true but am unable to find a counterexample. 
 A: In general the answer is no, even $F$ is a line bundle itself. It is easy to see that a globally generated line bundle is nef, and if $F$ is not nef, and the segment between $F$ and $M$ does not intersect with the ample cone in $N^{1}(X)$, then $F \otimes M^{n}$ is numerically propotional to a divisor lies in the interior of the segment, thus is not nef.
A: Here is a simple counterexample (of the form Zhengyu Hu suggested). Take $X$ to be the blowup of a point in $\mathbb P^2$, $M$ the pullback of $\mathcal O(1)$ under the blowup map, and $F$ the line bundle associated to the exceptional divisor $E$. Sections of $F\otimes M^n$ are rational functions which may have a pole of order 1 along $E$ and a pole of order up to $n$ along a line not meeting $E$. However, any rational function $f$ actually having a pole along $E$ (i.e. generating $F\otimes M^n$ at the points of $E$) must have a pole along some other divisor passing through $E$ (since $X$ has the same rational functions as $\mathbb P^2$).
A: In my view being big corresponds to being "generically ample", that is being ample outside a closed subset. This closed subset is called the augmented base locus of $M$ and it is often denoted by $\mathbb{B}_+(M)$ ( see http://de.arxiv.org/abs/math/0308116v2.pdf).
So, if for example you take $M$ big and $F$ to be a line bundle you can prove that $M^m\otimes F$ is always globally generated (i.e. base point free) outside the augmented base locus of $M$ (note that you don't need global generation of $M$ for this).
In the example of Anton Geraschenko in fact $\mathbb{B}_+(M)$ is exactly the exceptional divisor.
Something similar might hold if you consider locally free sheaves of higher rank.
