
mechanics question
i've already tried this and i get the answers to be a) 1.49 and b) 1.01 (both to 2dp) but i think these are both wrong.
a particle of mass 1.5kg rests in equilibrium on a rough plane under the action of a force of magnitude XN(ewtons) acting up a line of greatest slope of the plane. the plane is inclined at 25 degrees to the horizontal. the particle is in limited equilibrium and on the point of moving up the plane. the coefficient of friction between the particle and the plane is 0.25.
a) calculate the normal reaction of the plane on the particle
b) calculate the value of X

Hello djr8793
The forces acting on the particle are: $\displaystyle \vec F$, the weight $\displaystyle m\vec g$, the reaction of the plane $\displaystyle \vec N$, and the friction $\displaystyle \vec f$.
We have: $\displaystyle \vec F + m \vec g + \vec N + \vec f = \vec 0$.
If we project this on an axis parallel to $\displaystyle \vec N$, we get: $\displaystyle N  mg.cos(25) = 0$, therefore: $\displaystyle N = 15cos(25) = 13.6N$.
Now, you can find the value of the friction: $\displaystyle f = \mu N$.
If we project above relation on an axis parallel to $\displaystyle \vec F$, we get: $\displaystyle X  mg.sin(25)  f = 0$, therefore: $\displaystyle X = 9.74N$.
Hope this helps.