Acids: arachidic acid, palmitic acid, lauric acid, etc.
Alcohols: butan-1-ol, pentan-1-ol, etc.
The melting points of acids with $\ce{-COOH}$ at the end compared to melting points of alcohols, $\ce{-OH}$ at the end, is higher. I need to relate this to intermolecular forces. I was going to say that because of many more electrons in the acids, there are greater London dispersion forces and also more polarity (dipole dipole forces) because of the $\ce{COOH}$ at one end. Is this correct?
The ultimate goal of melting-point reasoning is to explain how intermolecular forces lead to better packing of molecules in the solid phase.
Both alcohols and organic acids have hydrogen bonds, but organic acids have stronger hydrogen bonds due to the withdrawing carbonyl group, $\ce{COO-}$. This causes the acids to interact stronger with each other. More importantly, organic acids form dimers (usually) with their hydrogen bonds (acetic acid dimer, for example). For this reason, the packing of the acids in the solid phase is significantly better than that of alcohol; hence the higher melting point. Acetic acid: $16~^\circ\mathrm{C}$ vs. $-114~^\circ\mathrm{C}$ for ethanol.
Huy Nguyen has already addressed the importance of dimers through hydrogen bonds in the solid state of alkanoic acids.
If you want to correlate the chain length of alkanoic acids of the form $\mathrm{{C_{n}H_{{2n+1}}-COOH}}$ with their melting points, a look at the data might be helpful:
\begin{array}{lrr} \mathrm{(trivial)\ name} & \mathrm{n} & \mathrm{mp}\\ \mathrm{\tilde{} acid} & & \mathrm{[°C]}\\ \hline \mathrm{acetic} & 1 & 17\\ \mathrm{propionic} & 2 & -21\\ \mathrm{butyric} & 3 & -5\\ \mathrm{valeric} & 4 & -32\\ \mathrm{caproic} & 5 & -3.4\\ \mathrm{enathic} & 6 & -7.5\\ \mathrm{caprylic} & 7 & 16.7\\ \mathrm{pelargonic} & 8 & 12.5\\ \mathrm{capric} & 9 & 31.6\\ \mathrm{undecanoic} & 10 & 28.6\\ \mathrm{lauric} & 11 & 48.3\\ \mathrm{tridecanoic} & 12 & 41.5\\ \mathrm{myristic} & 13 & 54.4\\ \mathrm{pentadecanoic} & 14 & 31.6\\ \mathrm{palmitic} & 15 & 28.6\\ \mathrm{margarinic} & 16 & 48.3\\ \mathrm{stearic} & 17 & 41.5\\ \mathrm{nonadecanoic} & 18 & 54.4\\ \mathrm{arachic} & 19 & 75.5\\ \mathrm{behemic} & 21 & 80\\ \mathrm{lignoceric} & 23 & 84.2\\ \mathrm{montanic} & 27 & 90.9\\ \end{array} (Source: German and english wikipedia)
Plotting melting points vs length of the alkyl substituent gives the following result:
