Answer:
Moment about SHOULDER ∑ τ = 3.17 N / m,
Moment respect to ELBOW Στ= 2.80 N m
Explanation:
For this exercise we can use Newton's second law relationships for rotational motion
∑ τ = I α
The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.
They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved
The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm
Moment about SHOULDER
∑ τ = I α
I = I_forearm + I_sphere
the forearm can be approximated as a fixed bar at one end
I_forearm = ⅓ m L²
the moment of inertia of the mass in the hand, let's approach as punctual
I_mass = m L²
we substitute
∑ τ = (⅓ m L² + M L²) α
let's calculate
∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10
∑ τ = 3.17 N / m
Moment with respect to ELBOW
In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow
Στ = I α
I = I_ forearm + I_mass
I = ⅓ m (L-0.03)² + M (L-0.03)²
let's calculate
i = ⅓ 2.3 0.47² + 0.5 0.47²
I = 0.2798 Kg m²
Στ = 0.2798 10
Στ= 2.80 N m
