ANSWER
122.5N per hand
EXPLANATION
First, we have to draw a free body diagram,
A wheelbarrow is an example of a lever but in this case, both forces are on the same side of the pivot - which is the wheel. To keep the wheelbarrow in balance, the output force, Fo, must be equal to the weight of the wheelbarrow,
[tex]F_o=m\cdot g=100\operatorname{kg}\cdot9.8m/s^2=980N[/tex]
Using the mechanical advantage relation between the input and output forces and the lever arms,
[tex]\frac{F_o}{F_i}=\frac{l_i}{l_o}_{}[/tex]
We have to find Fi. To solve for Fi, first raise both sides to the exponent of -1 - in other words, flip both fractions,
[tex]\frac{F_i}{F_o}=\frac{l_o}{l_i}[/tex]
And multiply both sides by Fo,
[tex]F_i=F_o\cdot\frac{l_o}{l_i}[/tex]
The output force is the one found before, 980N. Then, the length of the lever arm of the output force is the distance from the center of gravity to the wheel, 0.5m, and the length of the lever arm of the input force is the distance from the man's hands to the wheel, 2m,
[tex]F_i=980N\cdot\frac{0.5m}{2m}=\frac{980N}{4}=245N[/tex]
The total force the man has to make is 245N. Assuming each hand can support the same weight, then each hand should be doing 122.5N of force to keep the wheelbarrow in balance.
