The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude w), the normal force (mag. n), and the friction force (maximum mag. f ).
In the horizontal direction, we have
n cos(120º) + f cos(30º) = 0
-1/2 n + √3/2 f = 0
n = √3 f
and in the vertical,
n sin(120º) + f sin(30º) + (-w) = 0
n sin(120º) + f sin(30º) = (50 kg) (9.80 m/s²)
√3/2 n + 1/2 f = 490 N
Substitute n = √3 f and solve for f :
√3/2 (√3 f ) + 1/2 f = 490 N
2 f = 490 N
f = 245 N
(pointed up the incline)
