Question

based on the free - body diagram, the net force acting on this wheelbarrow is \boxed{} n.
f_y = 800 n (upward), f_y = - 800 n (downward), f_x = - 400 n (leftward), f_x = 250 n (rightward)

Explanation:

Step1: Calculate net force in y - direction

Forces in y - direction: \(F_{y1}=800\ N\) (upward) and \(F_{y2}=- 800\ N\) (downward, negative as it's opposite to upward). The net force in y - direction \(F_{net - y}=F_{y1}+F_{y2}=800+( - 800)=0\ N\).

Step2: Calculate net force in x - direction

Forces in x - direction: \(F_{x1}=-400\ N\) (leftward, negative as it's opposite to rightward) and \(F_{x2}=250\ N\) (rightward). The net force in x - direction \(F_{net - x}=F_{x1}+F_{x2}=-400 + 250=-150\ N\). The magnitude of the net force is \(|F_{net - x}| = 150\ N\) (since net force in y - direction is 0, the total net force is equal to the net force in x - direction).

Answer:

150