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.)

Explanation:

Step1: Calculate the weight of the milk carton

The weight of an object is given by \(F = mg\), where \(m = 1.76\space kg\) and \(g=9.8\space m/s^{2}\). So, \(F_{r}=mg=1.76\times9.8 = 17.248\space N\)

Step2: Use the torque equilibrium equation \(\sum\tau = 0\)

Torque \(\tau=rF\sin\theta\). Let the distance of the force \(F_{r}\) from the pivot be \(r_{1}=25.0 + 8.00=33.0\space cm = 0.33\space m\) (assuming the pivot is at the elbow - like joint, and the perpendicular distance for \(F_{r}\) is the total arm - length from the pivot to the hand). The distance of the biceps force \(F_{b}\) from the pivot is \(r_{2}=8.00\space cm=0.08\space m\), and \(\theta = 75.0^{\circ}\).

Since \(\sum\tau = F_{b}r_{2}\sin75^{\circ}-F_{r}r_{1}=0\) (counter - clockwise torque \(=\) clockwise torque), we can solve for \(F_{b}\).

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Substitute \(F_{r} = 17.248\space N\), \(r_{1}=0.33\space m\), \(r_{2}=0.08\space m\), and \(\sin75^{\circ}\approx0.966\)

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Answer:

\(73.7\space N\)