The problem reads: Two forces act on a 5.0 kg block on a friction-less surface.
(a) Draw a free-body diagram
(b) Determine the magnitude of the normal force (Fn)
(c) Determine net horizontal force
(d) Determine the magnitude and direction of the horizontal acceleration
I don't understand how to calculate the direction of the horizontal acceleration...usually for direction I am used to using
θ = arctan (∑FY/∑FX), however for this problem do I just completely ignore the Y net forces since it is asking for horizontal? Therefore is it just θ = arctan (∑Fx)?
I was wondering if anyone could also please check my work? Criticism and tips for my free-body diagram would be much appreciated because I struggle with those.
(b) For determining the normal force because I assume acceleration is 0 in the vertical position
∑Fy = Fn - mg sin10° = 0
Fn = mg sin10°
Fn = (5 kg)(9.80 m/s^2) sin10° = 8.5 N
(c) For determing the net horizontal force I calculated the forces F1 and F2 and added them up
F1: 50 N cos10° = 49 N
F2: 15 N cos 180° = -15 N
∑Fx = 49 N + (-15 N) = 34 N
I'm not quite sure why I had to calculate F1 and F2 forces using trigonometry when they're already given in the problem...however, I followed this step from a similar homework problem that required that. I'm not sure why this step is required, I merely assumed that's what I had to do so I feel iffy about this part.
(d) For determining the magnitude of the horizontal acceleration I used ∑Fx = m ax and solved for ax = ∑Fx / m
ax = (34 N) / (5 kg) = 6.8 m/s^2
For determining the direction of the horizontal acceleration I'm not sure what to do....
I suppose it's θ = arctan (∑Fx) ? I really don't know, any help would be much appreciated. Also I suppose the direction is NW?
Thanks for taking the time to help out.