HOMEWORK HELPER FOR INCLINED PLANE PROBLEMS

 


Step 1)  Draw the free-body diagram and put in an x-y coordinate system, with the x-axis parallel to the inclined plane:

 

 

 

 

 

If you know which direction the object will move, call that direction positive.  In this problem, I don't know which way the block will move, if at all.  So I will call up the ramp positive. 

Step 2) Draw all forces on the diagram.  We must add Fg , Ffriction and FN.  The force of friction will always oppose motion.  I know the applied force is greater than the force of gravity pulling the block down the incline, so friction (whether static or kinetic) will point down the ramp.

 

 

 

 

 

 

 

Step 3) Identify all forces not parallel to the x- or y-axis.  Resolve each such force into its x and y component, and then cross out that force.  In this problem the only force not parallel to the x or y-axes is the force of gravity.

Step 4) Resolving the force of gravity: The difference here is that the y-component is adjacent to the angle that we will use, so Fgy= Fg cos θ, instead of sin θ.

 

 

 

 

 

 

 

 

Now you can reposition the x-component of gravity so that you see it pulling the block down the incline:

 

 

 

 

 

 

 

Step 5) Now you are ready to apply the Mother of All equations: FNet = ma:

a) First work in the y-direction:  We know the object is not rising or falling in the y-direction, so ay = 0, thus FNety = 0.  We know that net force means the sum of all forces, so we add up all forces in the y-direction and set them equal to zero.  There are only two y-direction forces, FN points up so it is positive, and the y-component of gravity (Fgy) which points down so it is negative:

FN + - 16.99 N = 0, or FN - 16.99 N =0.  Solving for FN, we get FN = 16.99 N.

b) Next work in the x-direction: This is the difficult part.  Because there is friction, we must determine whether the friction is static or kinetic.  We can only do this by answering the musical question, "Will it move?"

To answer this question, we must first find the maximum static friction.  Here, the coefficient of static friction is 0.74 and the normal force is 16.99 N.  So the maximum static friction is 0.74*16.99 N = 12.6 N.  This means the block will not move unless the net force pulling on the block is greater than 12.6 N.  Let's look at our free body diagram again, but let's concentrate only on the x-direction forces:

 

 

 

 

 

 

 

Now, let's add the applied force of 20 N with the x-component of gravity (-9.81 N) to get a net force in the x-direction (without counting friction) of 10.2 N.  Let's redraw our FBD and replace the two force vectors (+20 N and -9.81 N) with one net force vector (+10.2 N) in the x-direction:

 

 

 

 

 

 

 

 

Remember that the maximum static friction is 12.6 N.  The net force pulling up the incline does not exceed that maximum.  Therefore, static friction will hold the block in place.  The block's acceleration is thus zero.  That means the x-direction forces are balanced, so the force of friction is also 10.2 N.  The final FBD looks like this: