\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\)
Chapter 1: Equilibrium
4.

Our First Physics Problem

If you said 40N, then not only were you right, but you just solved your first physics problem! (I hope it was easy.)

Sometimes physics problems can be as straightforward as this one, but most of the time trying to keep track of everything in your head is hard and can lead to errors, so it is useful to have a more formal system.

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The first step is to identify what we know and what we need. In this case, we know the tension in the rope and the weight of the block, and we don't know the normal force. We give each of these quantities a variable as a label and write them down. (I'm using ? to signify that this is what we want to solve for.) Because the block is in static equilibrium, we can write $$N+T=W$$ with the upward forces on the left and the downward forces on the right. Then we substitute in what we know. $$N+10=50$$ and finally we can solve for N by subtracting 10 from both sides. In a simple problem like this you don't have to be this formal, but it can be helpful in keeping you from making sloppy mistakes (like adding 10 and 50 together). Until you build confidence, sticking with a more formal method is the best bet.