The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Question about forces

Scroll to see replies

??
I've told you twice why the answer is mg sin a. Your answer is wrong, because it does not account for the fact that the weight is attached to the system. (And if it wasn't attached, it would fall off).
Reply 22
Avatar for james.h
james.h
11
Whispering Eye
why is the force not mg/sina ?

sina = mg/F no?

I think you're constructing your right-angled triangle wrong. :sadnod:
Think about where the angle aa actually is. Remember the whole system is tilted at an angle; perhaps consider rotating your coordinate system to suit, as you would for motion on a slope.

Either way, you get the gravitational force, mgmg, along the hypotenuse, and the force FF along the opposite side to the angle. Then basic trigonometry gives you:

sin⁡(a)=Fmg  ⟹  F=mgsin⁡(a)\displaystyle \sin(a) = \frac{F}{mg} \implies F = mg\sin(a) :yep:

You can apply a rudimentary check to this: does it make sense?
Well, as a→0, F→0; and as a→90∘, F→Fmaxa\to 0,\ F\to 0;\ \mathrm{and\ as\ }a\to 90^{\circ},\ F\to F_{max}. This is what you would intuitively expect, isn't it? When you apply a force perpendicular to the beam, it'll take the full 'push' from said force, but applying the same force along the length of the beam will have no net effect. :smile:

Related discussions

Latest

Trending

Trending

Articles for you