Why does the friction $-cv$ in a damped harmonic oscillator have a negative sign? 
Why is it that the friction force has a negative sign? Shouldn't it be positive as it is opposing the restoring force of the spring?
 A: The friction force in this example is the drag force due to the liquid, which resists to the motion of the objects, so it will be in the opposite direction of the velocity vector.
A: NB:The direction of the friction force is opposing the direction of motion, not the restoring force.
So, in this situation, when the body moves downward, the direction of the friction force is same with the restoring force. When the body moves upward,the direction of friction force is positive and is opposing the restoring force of the spring.
A: (Disclaimer: I have tried not to be too formal)
So, let's think about what we want our physical model to do. We start from the simple harmonic oscillator:
$$m\ddot{x} = \vec{F}=-kx.$$
This says that the acceleration of the mass becomes larger when the mass goes farther away from the equilibrium position, and at the same time is directed towards that equilibrium position (the minus in front of $k$). This produces an endless harmonic motion, where the mass is going back and forth around the equilibrium position.
Now, say that we doesn't like endless motions, because we have a real problem at hand. We want to have something that pushes our mass to equilibrium and holds it there. Or, in other words, we want something that slows down the oscillation at all times. How do we do this?
We introduce another force (the damping force), which intended to reduce the acceleration of the mass when it is positive and make it larger when it is negative. To make our lives easier, we define this force as proportional to the velocity of the mass, but in the opposite direction, and get:
$$m\ddot{x}=-kx-c\dot{x}.$$
Now, depending on the position of the mass, and of its velocity's direction, we have four cases. You can refer to the below (really quick) sketch of the situation. In it, our mass moves in the $y$ direction; the time is not shown; the $x$ direction is only for convenience. The equilibrium position is at $(0,0)$.

*

*The mass is above equilibrium and is moving upwards. I.e., $y>0$ and $\dot{y}>0$. This means that the damping force will be directed downwards, towards equilibrium.

*The mass is above equilibrium but is moving downwards. I.e., $y>0$ and $\dot{y}<0$. This means that the damping force will be directed upwards, away from equilibrium.

*The mass is below equilibrium and is moving upwards. I.e., $y<0$ and $\dot{y}>0$. This means that the damping force will be directed downwards.

*The mass is below equilibrium and is moving downwards. I.e., $y<0$ and $\dot{y}<0$. This means that the damping force will be directed upwards.

This is what we expect – a force that is acting against the oscillation direction.

A: Do you mean why is the viscous damping negative?
A spring reacts linearly to movement according to its spring constant, so when it shows x extending downwards there's a $F_{spring} =-kx $ reaction that goes the opposite direction.
The bottom also has what is called a damper.  In harmonic motion a damper is a component that acts linearly with velocity according to its dampening constant (c).  It isn't the spring itself resisting the motion. They also labelled the velocity as going in the down direction.  The force from a damper is then $F_{damper} = -cv $.  Since the motion and the change in motion are both the same direction, the reaction forces will be going the same direction opposite to the movement and velocity.
This might be a bit more detail than you need if you didn't realize that the reaction -CV is viscous damping, not an internal spring friction.
