\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\)

The Second Dimension

Now consider this example: a box sitting on the floor. There are four forces acting on the box: a downward force (weight, probably) of 10N, an upward force of \(N_1\), a rightward force of 25N, and a leftward force of \(N_2\).

You might think that this is very similar to the bridge example, and that we aren't given enough information to solve for the two unknowns. But actually we can show that \(N_1=10\u{N}\) and \(N_2=25\u{N}\). This is due to a very important principle which we will use a lot:

Thus there are actually two equations hidden in this example, not just one:

• the horizontal equation \(25=N_2\), and
• the vertical equation \(10=N_1\),

and with these two equations let us solve for the two unknowns.

If you have a force which is not horizontal or vertical, then you need to break up that force into its horizontal and vertical components, as discussed in Vector Basis and Notation. In this picture, the 20N force from the rope is at a 40° angle from the horizontal, and so it can be broken into a horizontal force with magnitude \(20\cos 40^\circ\) and a vertical force with magnitude \(20\sin 40^\circ\). The force balance equations are thus $$\begin{align} S&=20\cos 40^\circ \hbox{ (horizontal)}\\ N_T+20\sin 40^\circ &=50 \hbox{ (vertical)}\\ \end{align}$$