# Giancoli Solutions on Video

Learn physics easily with guided practice.

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## Features

• 1,930 video solutions for all regular problems in Giancoli's 7th Edition and 1,681 solutions for most regular problems in the 6th Edition.
• Final answer provided in text form for quick reference above each video, and formatted nicely as an equation, like $E=mc^2$. This is useful if you are in the library or have a slow internet connection.

• Pen colors make the step-by-step solutions clear. Red is used to illustrate algebra steps, and to substitute numeric values in the final step of a solution. When a solution switches to a new train of thought a different pen color emphasizes the switch, so that solutions are very methodical and organized.
• Solutions are classroom tested, and created by an experienced physics teacher.
• Videos are delivered with a high performance content delivery network. No waiting for videos to load or buffer.
• Pause, rewind, repeat, and never miss what is being said.

## Sample solution

Giancoli 7th Edition, Chapter 4, Problem 62

(1:36)

### Giancoli 6th Edition, Chapter 2, Problem 42

A LifeSaver!

Hi vtocco57, I've posted solutions for only the regular problems, but not the misconception quizes. Hopefully the many regular problem solutions will be enough for most subscribers.

All the best,
Mr. Dychko

Hi sugarhuny20, this is a common question. In the end, the answer depends on personal preference and which convention you choose to follow. Typically $g$ is taken to be the magnitude of the acceleration of gravity, which means it is always a positive number. Both I, and the textbook, follow that convention. That means the acceleration due to gravity is $-g$ if you have the typical coordinate system with down as the negative direction.
In the video for this solution the force of gravity was taken to be negative as you would expect, since it is directed down. That's the $F_{net} = F_T - F_g$ equation I'm referring to there. Notice the negative in front of the force due to gravity term. That negative takes care of the gravity's direction, so there is no need to introduce another negative when substituting for $g$.