A uniform beam XY is 100 cm long and weighs 4.0N. The beam rests on a pivot 60 cm from end X. A load of 8.0 N hangs from the beam 10 cm from end X. The beam is kept balanced by a force F acting on the beam 80 cm from end X. What is the magnitude of force F ? A 8.0N B 18N C 22N D 44N. Show full working.

C. 22N

Explanation: Note that we are considering a uniform beam XY which implies that it will be evenly divided. That is the weight will be in between.

One of the laws of equilibrium states that the sum forces across the clockwise moment = sum of forces across the anticlockwise moment

Hence from the diagram we have

(8×50)+(4×10) = 20 × X

400+40 = 20X

440 = 20X

X = 440/20

X = 22N

Hence the answer is option C which is 22N

c) up

hope this

the answer is a. hope this .