Question

a cook holds a 1.76 - kg carton of milk at arms length (see the figure below). what force \\( \vec { f } _ { b } \\) must be exerted by the biceps muscle? (ignore the weight of the forearm. give the magnitude.)
your response differs from the correct answer by more than \\( 10 \\% \\). double check your calculations. n

1. - / 1 points

a meter stick is found to balance at the \\( 49.7 - \mathrm { cm } \\) mark when placed on a fulcrum. when a 43.5 - gram mass is attached at the \\( 26.5 - \mathrm { cm } \\) mark, the fulcrum must be moved to the \\( 39.2 - \mathrm { cm } \\) mark for balance. what is the mass of the meter stick?

Explanation:

Step1: Calculate the torque due to the milk carton

The torque \(\tau_g\) due to the weight of the milk carton is given by \(\tau_g = r_gF_g\), where \(F_g=mg\) ( \(m = 1.76\space kg\), \(g = 9.8\space m/s^2\)) and \(r_g=(25.0 + 8.00)\space cm=0.33\space m\). So, \(F_g=1.76\times9.8 = 17.248\space N\) and \(\tau_g=0.33\times17.248\)

Step2: Calculate the torque due to the biceps muscle

The torque \(\tau_b\) due to the biceps muscle is \(\tau_b=r_bF_b\sin\theta\). The angle \(\theta = 75.0^{\circ}\), \(r_b = 8.00\space cm=0.08\space m\)

Step3: Set up the equilibrium equation

Since the arm is in rotational equilibrium \(\tau_g=\tau_b\). So, \(0.33\times17.248=0.08\times F_b\times\sin(75^{\circ})\)

$$F_b=\frac{0.33\times17.248}{0.08\times\sin(75^{\circ})}$$

$$F_b=\frac{5.69184}{0.08\times0.9659}$$

$$F_b=\frac{5.69184}{0.077272}$$

$$F_b\approx73.7\space N$$

Answer:

\(73.7\space N\)