Example 2

In this example, we have an object on a sloping frictionless surface, that is being pushed by an external force horizontally and we want to determine what will happen. Will the block accelerate up the ramp? Down the ramp? Not move at all?

(NOTE: the normal force is represented using the lowercase letter n here since that's what our previous textbook (and many other textbooks for that matter) used. Giancoli uses F_N.)

Step 1 : Account for all the forces acting on the object

Unless otherwise noted, we can infer that this object is on the Earth, so we know that gravity is acting downward resulting in a force of `mg' (in this case 980 N) `down'. The object is sitting on a surface that it can't pass through, so the surface (floor, ground, table-top, etc) is `pushing back' with a force called the normal force which acts perpendicular to the surface and directed away from it. Adding these two force vectors to the figure results in:

Step 2 : Define an appropriate coordinate system

Looking at the problem, we can see that the object is constrained to move along the incline (if it moves at all) so we'll define our +X axis to be in that direction, with +Y perpendicular to X and pointing up (well, our rotated up...) away from the ramp.

Step 3 : Convert all vectors into this new coordinate system

This is the most critical step. We know that the sum of all the vector forces acting on the object is equal to the mass times the vector acceleration the object will undergo. This is a 2-D (or depending on the situation, 3-D) vector equation which might be fairly complicated to solve. But we know the block here is not going to be moving arbitrarily around in 3-D space: it is just going to be moving along the incline in a line. That means it is really just 1-D motion, so we would like to come up with the F=ma equation in that direction alone. We also know that we can replace a vector by any two other vectors that add up to equal it, so this step involves replacing every force vector acting on the object with equivalent vectors that are aligned along the coordinate axes we chose. The first step here is to draw a little rectangle whose sides are parallel to the coordinate axes we defined, and which just surrounds the vector we are converting. The box should be placed so that the vector starts in one corner and ends in the opposite corner. One side of the rectangle that is aligned with the X axis gives us the component of the force in the X direction, and one side aligned with the Y axis gives us the component of the force in the Y direction. Now we can remove the original force (shown below as a dotted line) and replace it with these two vectors. The arrow heads should be placed so that these two vectors add up to the original one. We have now replaced the original force with two forces (one in each axis direction). The physics is completely unchanged by doing this: whether the object is being acted on by the single original force or these two new forces, it will behave in exactly the same way.

Step 4 : Solve for the magnitudes of the components

We need to solve for the magnitudes of these forces now. There is no general rule about whether sine or cosine or tangent or whatever might be used here. You'll need to look at the rectangle you drew and what angles you know or can infer and decide which trigonometric function you can use. In this case, the two blue lines are parallel, and the two red lines are parallel, so we can see that the angle in our triangle is the same 20 degree angle of the incline:

We can now replace the original force with these two forces:

and look at the triangle and what we know about it to find the magnitudes of those forces:

Step 5 : Replace the original force with the components we just found

Now we can go back to our original figure and replace the 1000 N force at 20 degrees, by the two forces we just computed:

Step 6 : Repeat for all other forces acting on the object

In this particular example, the weight of the object (which is a force down towards the center of the earth) is NOT lined up with our new coordinate system, so we need to convert it as well. We put the rectangle around this force vector (again making sure that the sides of the rectangle are in line with our new coordinate axis directions):

It may take a couple of steps to propagate the angles we know so we can end up with the angle we need to solve for the X and Y components of this force:

Finally ending up with the triangle we can use to solve for the magnitudes of the components:

Now we have finally converted all the forces into components that are lined up with our coordinate axes:

Step 7 : Solve F=ma in each direction

Now we can take the original F=ma vector equation and look at it separately in each direction.

Y direction : ΣF_y = ma_y

In this example, there is no friction, so we don't need to calculate the normal force, but if friction were present, we could find it now: in the Y direction, we have `n' pointing in the +Y direction, 342.0N in the -Y direction (i.e. part of the 1000N external force we applied is acting in that direction) and 920.9 N in the -Y direction, and we know that the block cannot accelerate in that direction, so ΣF_y = ma_y becomes +n-342N-921N=0 or n=921N+342N=1263N.

(If friction were present in the problem, we could find it now since it depends on the normal force, but we have no friction here so don't really end up using F_N at all in this version of the problem.)

X direction : ΣF_x = ma_x

Looking in the X direction now, we have +939.7N (coming from the X component of our original external force) and -335.2N (coming from the component of the weight in this rotated X direction) so ΣF_x = ma_x becomes: +939.7N - 335.2N = (100 kg)(a) or 604.5N = (100kg)(a) or a = 6.045 m/s^2.

(This was positive and our +X axis was pointing up the ramp, so apparently we're providing enough force to accelerate the object UP the incline.)