Uday M. answered • 09/27/19
M.S. Engineering, 5+ years of teaching experience
First, I would recommend drawing a free body diagram and accounting for all the forces.
Just to make the math easier, it's wise to change our coordinate system. Instead of the conventional horizontal x-axis and vertical y-axis, we can choose our "x" axis to be along the incline of the slope and the "y" axis to be perpendicular to that (normal to the slope). The reason we have chosen this is because we are trying to find the coefficient of static friction, and the static friction is 'pulling' the box up the incline.(our new "x" axis). Remember that the new coordinate system we choose must be perpendicular so that we can take advantage of the fact that motion in a direction does not affect motion in the perpendicular direction.
Now let's account for our forces. We have gravity (G) which goes straight down, so we have to convert it into our new coordinate system (you can use trig for that if you draw out the angles and vectors). Gravity will be split into Gx and Gy using trigonometry (this is where the angle of incline becomes important). We also have the normal force N from the slope pushing back on the box, which acts in the new y-direction. There is the force that can keep the box from sliding; this is the frictional force SF that acts in the x-direction.
(Remember there are two types of friction: static friction which is what stops the box from moving in the first place (SF), or kinetic friction which slows the box after it is already moving (KF). When we discuss friction, if the object starts off from rest, we usually need to figure out whether the applied force (in this case, gravity) is enough to overcome SF. If it isn't, then the box would not move. If it is, the box will move and be hindered by KF. Here, they tell us that the box is experiencing SF, so we know it does not move.)
Recall the equation for friction. For both SF and KF, it is μ*N, where μ is the coefficient of friction and N is the normal force. The μ value depends on the material: some materials are more 'rough' than others and cause more friction (e.g. sandpaper vs ice). The μ for SF and KF are also different: μ(SF) is higher than μ(KF) because it is harder to get an object to move from rest than it is to keep it moving. This is what we're solving for.
So in order to get the friction, we first have to find N since it is in the equation. Let's write Newton's second law in the y-direction.
Gy + N = m * a(y)
The weight of the object is the gravity that it is experiencing (remember to convert it into our new coordinate system using trigonometry). The box is not moving in the new y-direction. Here you can solve for N. Make sure your signs are appropriate.
Now, write Newton's second law in the x-direction.
Gx + SF = m * a(x)
Recall that this is static friction, so the box is not moving in the new x-direction either. Here you can plug in the equation for SF (and then subsequently plug in the value for N that you just solved for) and plug in the adjusted gravity (again, using trigonometry). Make sure your signs are appropriate. Solve for μ, and that's your answer.
