Answer to Question #242643 in Physics for Dave

Question

Two blocks of masses M1=20kg and M2=10kg are supported by a uniform horizontal beam of length and mass 6kg. Calculate (1) the normal force exerted by the fulcrum on the beam if it is placed under the center of gravity (2) the distance in which the mass M2 must be placed in order for the system to be balance

Answer 1

(1) The normal force is

$$N=g(M_1+M_2+m)=353\text{ N}.$$

(2) Assume M₁ is at distance $a$ from the fulcrum. Then we have the following torque equilibrium equation:

$$M_1a+\frac {ma}{a+b}·\frac a2=M_2b+\frac{mb}{a+b}·\frac b2,\\\space\\ b=\frac{\sqrt{a^2(2M_2-M_1)^2+4a^2(2M_2+m)(2M_1+m)}+a(2M_2-2M_1)}{2(2M_2+m)}=\\\space\\ =a\frac{(\sqrt{6788}+20)}{92}=1.11a.$$