- #1
King
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The question is, "A particle of mass 10kg is at rest on a rough plane inclined at 30° to the horizontal. A horizontal force of magnitude 10N acts on the particle. a) Find the magnitude of the friction force on the particle."
URL TO IMAGE (I added the green to help you understand why I am confused): i209.photobucket.com/albums/bb180/newguyjb/forcesdiagram.jpg
The book provides the answer with the working, which is:
F+10cos30°=98sin30°
F=98sin30°-10cos30°
=40.3N
However, I don't understand how they have decided that it is 10cos30°. On the diagram I provided (which is the link) in green the area which is not in the diagram in the book, but which relates to why I am confused as to why it is 10cos30°. Could someone explain why it is 10cos30°.
Thanks for any help.
URL TO IMAGE (I added the green to help you understand why I am confused): i209.photobucket.com/albums/bb180/newguyjb/forcesdiagram.jpg
The book provides the answer with the working, which is:
F+10cos30°=98sin30°
F=98sin30°-10cos30°
=40.3N
However, I don't understand how they have decided that it is 10cos30°. On the diagram I provided (which is the link) in green the area which is not in the diagram in the book, but which relates to why I am confused as to why it is 10cos30°. Could someone explain why it is 10cos30°.
Thanks for any help.