Answer
Denoting the upward force exerted by the second worker as $F_{2},$ then application of Newton's second law in the vertical direction yields $F_{N}=m g-F_{2},$ which leads to
$$
f_{s, \max }=\mu_{s} F_{N}=\mu_{s}\left(m g-F_{2}\right)
$$
In order to move the crate, $F$ must satisfy the condition $$F>f_{s, \max }=\mu_{s}\left(m g-F_{2}\right)$$
or
$$
110 \mathrm{N}>(0.37)\left[(35 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)-F_{2}\right] \text { }
$$
The minimum value of $F_{2}$ that satisfies this inequality is a value slightly bigger than $45.7 \mathrm{N},$ so we express our answer as
$$F_{2, \min }=46 \mathrm{N} $$
Work Step by Step
Denoting the upward force exerted by the second worker as $F_{2},$ then application of Newton's second law in the vertical direction yields $F_{N}=m g-F_{2},$ which leads to
$$
f_{s, \max }=\mu_{s} F_{N}=\mu_{s}\left(m g-F_{2}\right)
$$
In order to move the crate, $F$ must satisfy the condition $$F>f_{s, \max }=\mu_{s}\left(m g-F_{2}\right)$$
or
$$
110 \mathrm{N}>(0.37)\left[(35 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)-F_{2}\right] \text { }
$$
The minimum value of $F_{2}$ that satisfies this inequality is a value slightly bigger than $45.7 \mathrm{N},$ so we express our answer as
$$F_{2, \min }=46 \mathrm{N} $$
