A boy and a man carry uniform rod of length l horizontally in such a way that the boy get 1/4of the load.if the boy is at one end of the rod the distance of the fan from the other end is
L÷3 ans
L÷3
In the last step L 2/6
summation of torque = 0
Fb ×L/2-Fm×(L/2-x)=0
W/4×L/2-3W/4z×(L/2-x)=0
W/4×L/2 = 3W/4z×(L/2-x)
L/2=3L/2-3x
3x=L
X=L/3
Hope it helps.....!!!!
Not nice
Fantastico!
Let W be the weight of rod. The center of gravity of the rod is at the center, l/2
summing moments about the center of the rod.
(w/4)*l/2-(3w/4)(x)=0 where x is from the center.
x= l/6, so from the other end, l/2-l/6= L(3/6-1/6) or the distance from the other end is 2/3 * L