You are Here: Home >< Maths

1. ??
2. 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).
3. (Original post by Whispering Eye)
why is the force not mg/sina ?

sina = mg/F no?
I think you're constructing your right-angled triangle wrong.
Think about where the angle 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, , along the hypotenuse, and the force along the opposite side to the angle. Then basic trigonometry gives you:

You can apply a rudimentary check to this: does it make sense?
Well, as . 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.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 9, 2010
Today on TSR

### Exam Jam 2018

Join thousands of students this half term

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams