Question

An experimental subject's morphometric characteristics are listed below. If she holds a 5 kg weight steady with her elbow joint flexed at a 90 degree angle using only her biceps muscle, how much force (in kg) must the biceps generate? Distance from the biceps insertion to the hand = 25 cm Distance from the elbow joint to the hand = 30 cm.
a. 50 kgb. 18 kgc. 6 kgd. 30 kge. 70 kg

Answer 1

Answer:

c. 6 kg

Explanation:

= torque due to weight of 5 kg about elbow joint

= 5 x 30 kg cm

torque due to force F by muscle about elbow joint  in kg

25 X F kg cm

for rotational equilibrium

25 X F = 5 x 30

F = 5 x 30 / 25

= 6 kg

Answer 2

Answer:

$E=58.8\ N=6\ kgf$ force must be generated by the biceps.

Explanation:

Given:

force at acting normal to the bent elbow hand, $F=5\ kgf=5\times 9.8\ N$

distance from the load on the hand to the elbow joint, $A=0.3\ m$

distance from the biceps to the hand, $B=0.25\ m$

Now in order to hold the load the total moment should be balanced.

So,

$F\times A=B\times E$

where:

E = effort generated by the biceps

$(5\times 9.8)\times 0.3=0.25\times E$

$E=58.8\ N=6\ kgf$ force must be generated by the biceps.