Thread: forces on an inclined plane

1. forces on an inclined plane

my problem has been resolved this far but i cant transpose or what ever you call it

rearrange the formula

ive been getting help and have ended up with this

24.5 - x.cos(30). = u(42.44 + x.sin(30))

where u is cofficient of friction 1/7 or 0.143

also should the brackets be around the 30??

30 is the angle X is at

so 5g is resolved but cant work out X
49cos30= 42.44
49sin30= 24.5
but i cannot work it out. sure you all know what its like after a day of exams.
and a break down of how to work it would be great

2. Use the unit circle to see that $\displaystyle \sin(30) = \dfrac{1}{2}$ and $\displaystyle \cos(30) = \dfrac{\sqrt{3}}{2}$ and you're given $\displaystyle \mu = \dfrac{1}{7}$

$\displaystyle \displaystyle 24.5 - \dfrac{x \sqrt{3}}{2} = \dfrac{1}{7} (28.3 + \frac{x}{2})$

1. it has to be more straight forward than that

2.i dont understand any of that

have a look and see if the equation i got is even right 5g i can resolve but i cant resolve X

4. solved it

how do show its solved or is this enough

5. I'm not sure what even is the question... are all those forces in equilibrium and you need to generate two equations showing the relationship of the forces?

If so, resolve along and perpendicular to the plane:

$\displaystyle X \cos(30) + F = 5g \sin(30)$

$\displaystyle 5g \cos(30) + X\sin(30) = R$