Applications of Dot Product (full lesson) | MCV4U

[@alyssapeterson3339]

Thank you for your amazing videos!

[@thedeathofbirth0763]

Thank you for a wonderfully explained lesson!

[@williamwelmans8648]

Thanks! A well explained tutorial.

[@krishmahadik3506]

I wish I had ur brain for my exams

[@baloolnamil9550]

Fr

[@maysunsaad1525]

Very helpful, Thanks . Can the work be negative (dot product is negative) what does it mean?

[@MrJensenMath10]

Yes it can. I just means the component of the force parallel to the direction of movement is in the opposite direction of the movement.

[@thedeathofbirth0763]

Jensen is there any possibility for you to explain HW #3 for this lesson, question number nine! Why the angle between the 120 N force on the ramp inclined at 15 degrees make the angle between 120 force applied at 35 degrees and the displacement of 12 meters to be 20 ? Please man, I won't be able to sleep until I get this. I wish I could be your student in your class....

[@MrJensenMath10]

The key is that the force for the second part is the exact same as the force in the first part of the question. Meaning it still makes an angle of 35 degrees with level ground. When you calculate the dot product you use the angle between the force and the displacement. Since the displacement is 15 degrees above level ground, and the force is 35 degrees above level ground, the angle between the two is 20 degrees. Hope that helps.

[@thedeathofbirth0763]

@@MrJensenMath10 Ohhh Thank you so very much! I am not as good as I thought! I am an idiot! I kept putting the tail of the force to the tail of the ramp and kept thinking my angel still is 35 degrees! How can I be so stupid when I have such a great teacher as you! Many thanks Jensen.... Have a great Monday!

[@thedeathofbirth0763]

Jesen, I just came back from work and there is no answer. I am really stuck on this question. All I could think of was this. Can you Or anyone who is on this site explain how cosine of 20 degrees was calculated? Thank you.