Question

A

cook
holds a 3.2 kg carton of milk at arm's length.
75.9
w
25,5 cm
What force FB must be exerted by the bi-
ceps muscle? The acceleration of gravity is
9.8 m/s2. (Ignore the weight of the forearm.)
Answer in units of N.

Answer 1

Step-by-step explanation:

From the given information:

From the rotational axis, the distance of the force of gravity is:

d_g = 25+5.0 cm

d_g = 30.0 cm

d_g = 30.0 × 10⁻² m

However, the relative distance of FB cos 75.9° from the axis is computed as:

d_B = 5.0 cm

d_B = 5.0 × 10⁻² m

The net torque rotational equilibrium = zero (0)

i.e.

`\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = (F_g d_g)/(F_g cos 65.6) \\ \\ F_B = ((3.2)(9.8)(30*10^(-2)))/((5.0*10^(-2) * cos 75.9)) \\ \\ \mathbf{F_B = 772.4 N}`

= 772.4 N

Thus, the force exerted = 1772.4 N