Hi eazyduzit,
My advice for solving this problem would be
1) Set up a Cartesian coordinate system that is aligned with the inclined plane. I.e. the \(\displaystyle x\) coordinate varies in the direction
along (parallel to) the surface of the plane, and the \(\displaystyle y\) coordinate varies in the direction perpendicular to the surface of the plane. You have to pick a sign convention too. Let's say that the \(\displaystyle +x\) direction is the direction moving up the incline, while sliding down the incline is decreasing \(\displaystyle x\). Let's say that the \(\displaystyle +y\) direction is moving away from the surface of the incline, while moving towards it decreases \(\displaystyle y\). Orienting the coordinate axes to line up with the incline is simply more convenient in this situation than choosing them to be horizontal and vertical (relative to the page).
2) Draw a
free body diagram of the box
. This is a diagram in which you consider the box in isolation, drawing only it, and the forces that act on it. That makes it much easier to see what is affecting the state of motion of the box, without being confused by extraneous details in the surroundings. In order to do this, you have to make a tally of all the forces that are acting on the box. As far as I can tell, these are:
- the box's weight (call this \(\displaystyle w\)). I.e. the force due to gravity on the box. As always, \(\displaystyle \vec{w} = m\vec{g}\), where m = 2 kg is the mass of the box, and \(\displaystyle \vec{g}\) = 9.81 N/kg, downward, is the acceleration due to gravity (on Earth's surface). The weight force vector acts vertically downward, which means that you will need to resolve it (i.e. decompose it) into \(\displaystyle x\) and \(\displaystyle y\) components \(\displaystyle w_x\) and \(\displaystyle w_y\)
- The force due to the tension in the rope (call this \(\displaystyle T\)). The tension force vector \(\displaystyle \vec{T}\) acts at an angle of \(\displaystyle \alpha\) above the surface of the plane, which means that you will need to resolve it into components \(\displaystyle T_x\) and \(\displaystyle T_y\).
- The force due to friction with the surface of the plane (call this \(\displaystyle f\)). This force acts directly in the \(\displaystyle x\) direction. If we assume that, in the absence of any forces other than gravity, the box would want to move in the \(\displaystyle -x\) direction (to slide down the plane), then the frictional force acts purely in the \(\displaystyle +x\) direction to oppose this motion. The magnitude of the friction force is given by \(\displaystyle f = \mu N\), where \(\displaystyle N\) is the normal force acting on the box (see below). The coefficient of friction (of static friction in this case), \(\displaystyle \mu\) is stated to be 1/3 for these two surfaces (box and plane).
- The normal force (\(\displaystyle N \)) on the box from the surface of the plane. This is a reaction force (the surface of the plane pushes on the box). The word "normal" means "perpendicular" in this context, and this is the force on the box that acts purely perpendicular to the surface of the plane i.e. purely in the \(\displaystyle +y\) direction.
3) The statement that the box is in equilibrium is a statement that there is
no net force acting on it. In other words, all the above listed force vectors, when summed up as vectors, add up to zero resultant force:
\(\displaystyle \displaystyle \sum \vec{F} = \vec{0} \)
We can break this sum up into two separate sums of the components of the forces in each of the \(\displaystyle x\) and \(\displaystyle y\) directions (where these components have signs):
\(\displaystyle \displaystyle \sum F_x = w_x + T_x + f = 0 \)
\(\displaystyle \displaystyle \sum F_y = w_y + T_y + N = 0 \)
Can you take it from here? You need to actually compute each of these force components in terms of known quantities.
Note that the procedure I've outlined above is the approach to take in solving
any statics problem, not just this one.