(4) Trying to pull a boulder up a hill

This gives us a practical application of using components of vector quantities.
The picture below shows the forces involved when a person tries to pull a boulder up a grassy slope.
forces acting on the boulder

The forces are all acting in different directions, and the person has not yet succeeded in moving the boulder. We therefore know from Newton's first law that there is no net resultant force acting on the boulder.
If I write the forces as w, f, n and p, then we have the vector equation
w + f + n + p = o

where o stands for zero force.
A very useful result of this is that the components of the four forces, w, f, n and p, acting in any particular chosen direction, must balance each other out. There is no net force acting in any specified direction on the boulder.
Also, since there is no sideways force acting on this boulder, all the forces are acting in two dimensions. This means that we can write each force in terms of any two non-parallel vectors which we choose, (components), and doing this will give us two independent equations of balanced forces, one for each chosen vector direction.
It makes things easier if we have at least one force acting entirely in one of the chosen directions.
The weight acts vertically so one possibility is to split all the forces into their horizontal and vertical components. I've done this in the picture below.
I've called the magnitudes of the forces w for the weight, f for the friction, n for the normal reaction of the ground and p for the pull in the rope.
resolving the forces horizontally and vertically

Effectively we are resolving the forces so that they are described in terms of the two unit vectors i and j and then saying that the net force in both the i and j directions must be zero.
Resolving horizontally, (considering the i components), we have
p cos (A+B) = n sin A + f cos A

and, resolving vertically, (considering the j components), we have
n cos A + p sin (A+B) = f sin A + w

We can then use these two equations to calculate the sizes of any two of these quantities provided we know the sizes of the rest.
The forces also split up neatly if we choose directions perpendicular and parallel to the slope of the hill.
I've done this in the diagram below.
resolving the forces parallel and perpendicular to the slope

Effectively we are now resolving the forces so that they are described in terms of the unit vectors s and t and then saying that the net force in both the s and t directions must be zero.
Resolving parallel to the slope, (considering the s components), we have
f + w sin A = p cos B

and resolving perpendicular to the slope, (considering the t components), we have
n + p sin B = w cos A

Again, we can use these two equations to calculate the sizes of any two of these quantities provided we know the sizes of the rest.

Also, if we wanted to, we could use two vectors not at right angles to get our two equations.
For example, if we chose j and s we would have, resolving vertically in the j direction,

n cos A + p sin (A+B) = f sin A + w

and, resolving parallel to the slope in the s direction,
f + w sin A = p cos B

Any non-parallel pair of vectors will work but we won't get any new information by including an equation in the direction of a third vector because every force in this diagram needs a combination of only two non-parallel vectors to describe it completely.

We can now use vector components to give us a way of actually finding the magnitude of a vector quantity.


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