A man carries a load of mass 2.6kg from one end of a uniform pole 100cm long which has a mass 0.4kg. The pole rest on his shoulder at a point 60.0cm from the load and he holds it by the other end. What vertical force must be applied by his hand and what is the force on his shoulder?

F = 39.2 N   (hand force) and    N = 68.6 N (shoulder force)

Explanation:

In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole

Let's set the reference system at the fit point of the shoulder

     ∑ τ = 0

We will assume that the counterclockwise turns are positive

    w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0

all distances are measured from the support of the man (x₀ = 0.60 m)

    F = (w₁ 0.60 + W 0.1) / 0.4

    F = (m₁ 0.6 + m 0.1) g / 0.4

let's calculate

    F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4

    F = 39.2 N

this is the force that the hand must exert to keep the system in balance

We apply the translational equilibrium condition

    -w₁ -W + N - F = 0

     N = w₁ + W + F

     N = (m₁ + m) g + F

let's calculate

     N = (2.6 + 0.4) 9.8 + 39.2

     N = 68.6 N

answer; (arch); good luck