A uniform bar 25 meters long weighs 10,000 Newton's. From end A weight of 2500 Newton's is hung. At B the other end of the bar, there is a weight of 3500 Newton's. An upward force of 3000 Newton's is exerted 4 meters from B, while and upward force of 4000 Newton's is exerted 8 meters from A. Determine the magnitude, direction and point of application of the force needed to establish transitional and rotational equilibrium.

dont know where to even start. please help

2 answers

force, direction, distance from A

F , up (assumed) , x (unknown)

2500 , down , 0

4000, up , 8

10,000, down , 12.5

3000, up , 21

3500 , down , 25

sum of forces up = 0
so
-2500+4000-10000+F+3000-3500 = 0
F -5500 = 0
so
F = +5500 up

sum of clockwise moment = 0 about A
2500*0 -4000*6 +10000*12.5 -F*x-3000*21 + 3500*25 = 0
-24,000 +125,000 -F*x -63,000 +87,500=0
F * x = 125,500 but F = 5500
so
x = 22.8 meters from A
hmm the answer is 9000N upward 12 meters from B?