I have written it down, the y direction would be the direction of Fn in this case.
But thanks for the help, I'll do it later. No need to reply after this.
T=m(a+gsin(8)-Us(gcos(8)) my bad.
In the y-direction there is a component of weight and also the normal force. positive would be uphill. so the component of weight is negative and the normal force would be positive. Now I get: Fny=m(a+gcos8) so here I would subtract the x component? adding...
a=Fnet/m I don't know where to go from here. I already broke into FBD and tried to solve it already. When I redid it I ended up with the same thing.
Oh, that g is a typo so it's not affecting my current calculation.
T=mg(a+gsin(8)-Us(gcos(8))
Homework Statement
A girl of mass mg = 52 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.157; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ a max = 0.015 m/s2...