Physics help: Zoltan supports a 5 m long 10 kg beam at one end. The beam is uniform in cross section and in...

Zoltan supports a 5 m long 10 kg beam at one end. The beam is uniform in cross section and in density, and is supported at the other end by a bench.

a) What conditions must be satisfied for a rigid body to be in static equilibrium?

(b) What force must Zoltan exert in order to keep the beam horizontal? Justify your answer.


Note: by answering, you are not doing my homework. I am studying for exams and this was a past exam question, thanks!

a) there are 2 conditions for any rigid body to be in equilibrium
i) Summation of all forces acting on any object must be zero
ii) Summation of all torques acting on any object must be zero OR total clockwise torque must be equal to total anticlockwise torque.

b) Here the forces acing on beam are:
i) weight of beam acting vertically downward which is equal to 10*9.8 = 98N (W = mg)
ii) F1, force applied by the bench on beam acting vertically upward
iii) F2, force applied by Zoltan on beam acting vertically upward

Since, the beam is uniform (it reflects that beam's center of gravity coincides with the geometrical center of beam) and is supported by its ends on both sides, it means the distance from beam's center of gravity to the acting point of either force is equal (d1 = d2). Therefore from 2nd condition of equilibrium
clockwise torque = anticlockwise torque
F1*d1 = F2*d2 (Since d1=d2=d)
F1*d = F2*d
F1 = F2 (d is cancelled by opposite sided d)
So, F1= F2 = F

Now, According to 1st condition of equilibrium; summation of all forces must be zero

F1 + F2 -W = 0 (Since F1 and F2 are in same direction therefore they are added and 'W' is in opposite direction so it is subtracted)
or F1 + F2 = W (send 'W' on opposite side)

But F1 = F2 = F (Have been proved above)
then
F + F = W
2F = 98 (W=98N)
or F = 49N