Question

 The wheel with the load weighs 120kg. The center of gravity of the wheel is at a distance of 40 cm from the axis of rotation. With what force do we lift the wheel if the end of the hands is at a distance of 1.2 m?

Answer 1

Gives

Mass =120 kg

Weight (w)=mg

$$W=120\times9.8=1176N$$

Distance of center of gravity=40cm=0.4m

Wheels end of the hand distance=1.2m

Transation equation

$$R_f+R_b=mg=120\times9.8=1176N$$

For rotation equation on taking the torque about C.G

$R_f(0.4)=R_b(1.2-0.4)$

$0.4R_f=0.8R_b$

$R_b=\frac{1}{2}R_f$

$R_b+R_f=1176N$

$\frac{3}{2}R_f=1176N$

$R_f=784N$

$R_b=392N$

Each wheel force

$R_f=\frac{784}{2}=392N$

$R_b=\frac{392}{2}=196N$