A 9.0 m long uniform rod with a weight of 8 N has 3 masses hanging from it; a 4N mass 1 meter from the left end, a 2 N mass 6 meters from the left end and a 3 N mass 4 meters from the left end. What force must be applied to bring the rod into total mechanical equilibrium

Explanation:

Assume the rod is horizontal

Sum vertical forces to zero

F - 8 - 4 - 2 - 3 = 0

F = 17 N

While this result answers precisely the question asked, it is not sufficient to guarantee equilibrium. We must also locate this force by summing moments to zero. Let's choose to sum about the left end.

8[9/2] + 4[1] + 2[6] + 3[4] - 17[x] = 0

x = 3.76470588...

the force must be supplied upward approximately 3.765 m from the left end

The rod does not actually have to be horizontal. Moment arms will all be proportionately reduced by tilting the bar from horizontal and the results will be identical. Assuming horizontal just reduces the calculation effort.

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.