Question

Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from the axis of his elbow. If his biceps brachii muscle pulls at a line 5 cm (0.05) from the axis of the elbow, how much force must it pull with in order to maintain an isometric muscle action (i.e. static equilibrium)?

Answer 1

Answer:

Force due to biceps is given as

$F = 1991.05 N$

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

$mg \times L = F \times d$

so we have

$18.14 \times 9.8 \times 0.56 = F \times (0.05)$

$F = 1991.05 N$